Explanation:
Okay, so I need to find the area of the set S, which consists of points (a, b) where 0 ≤ a, b ≤ 1, such that the equation x⁴ + a x³ - b x² + a x + 1 = 0 has at least one real root. Hmm, let me think about how to approach this.

First, the equation is a quartic polynomial. Quartic equations can be tricky, but maybe there's some symmetry or substitution that can simplify things. Let me write down the equation again:

x⁴ + a x³ - b x² + a x + 1 = 0.

Looking at the coefficients, I notice that the coefficients of x⁴ and the constant term are both 1, and the coefficients of x³ and x are both a. The middle term is -b x². Maybe there's some symmetry here. Let me check if the polynomial is palindromic or something like that.

A palindromic polynomial has coefficients that read the same forwards and backwards. Let me check:

Original coefficients: 1 (x⁴), a (x³), -b (x²), a (x), 1 (constant). So yes! It's a palindromic polynomial. That means if we reverse the coefficients, we get the same polynomial. For palindromic polynomials of even degree, there's a substitution that can be used. Specifically, for a palindromic polynomial of degree 4, we can divide both sides by x² to make it symmetric.

Let me try that. Dividing the equation by x² (assuming x ≠ 0, but we can check x=0 separately later):

x² + a x - b + a / x + 1 / x² = 0.

Now, let me set y = x + 1/x. Then, note that y² = x² + 2 + 1/x², so x² + 1/x² = y² - 2. Similarly, x³ + 1/x³ = y³ - 3y, but maybe I don't need that here.

Looking at the equation:

x² + 1/x² + a(x + 1/x) - b = 0.

Which is:

(y² - 2) + a y - b = 0.

So, simplifying:

y² + a y - (b + 2) = 0.

Therefore, the original equation can be transformed into a quadratic in terms of y. However, this substitution is valid only when x ≠ 0. But since we are looking for real roots, we need to check if x=0 is a root. Let me check x=0:

Plugging x=0 into the original equation: 0 + 0 - 0 + 0 + 1 = 1 ≠ 0. So x=0 is not a root. Therefore, all real roots satisfy x ≠ 0, so the substitution is valid for all real roots.

Therefore, the original equation has a real root if and only if the equation y² + a y - (b + 2) = 0 has a real solution y such that y = x + 1/x for some real x ≠ 0.

But we need to relate this back to real x. So, for each real solution y of the quadratic equation, we need to check if there exists a real x ≠ 0 such that y = x + 1/x.

So, first, let's analyze the quadratic equation in y:

y² + a y - (b + 2) = 0.

Let me denote this as:

y² + a y - (b + 2) = 0.

The solutions for y are:

y = [-a ± √(a² + 4(b + 2))]/2.

But for real solutions y, the discriminant must be non-negative. However, even if the discriminant is non-negative, we need to check if the corresponding y can be expressed as x + 1/x for some real x.

So, first, the discriminant D = a² + 4(b + 2) must be ≥ 0. But since a and b are between 0 and 1, let's check:

a² is between 0 and 1, and 4(b + 2) is between 4(0 + 2) = 8 and 4(1 + 2) = 12. Therefore, D is always positive. So the quadratic equation in y always has two real roots. Therefore, the original equation will have real roots if and only if at least one of these y's can be expressed as x + 1/x for some real x ≠ 0.

So, now, we need to find for which (a, b) there exists a real number y such that y = x + 1/x and y satisfies the quadratic equation above.

But what is the range of possible y = x + 1/x for real x ≠ 0?

Let me recall that for real x ≠ 0, the expression x + 1/x has a range of (-∞, -2] ∪ [2, ∞). This is because, for x > 0, by AM ≥ GM, x + 1/x ≥ 2, and for x < 0, let me set x = -t where t > 0, then x + 1/x = -t - 1/t = - (t + 1/t) ≤ -2. Therefore, the range of y is (-∞, -2] ∪ [2, ∞).

Therefore, for the original equation to have a real root, the quadratic equation in y must have at least one root in (-∞, -2] ∪ [2, ∞).

Therefore, the problem reduces to finding the set of (a, b) in [0,1]x[0,1] such that the quadratic equation y² + a y - (b + 2) = 0 has at least one root in (-∞, -2] ∪ [2, ∞).

So, now, we need to find the conditions on a and b such that this quadratic equation has at least one root in (-∞, -2] ∪ [2, ∞).

Let me denote the quadratic equation as:

y² + a y - (b + 2) = 0.

Let me denote this as f(y) = y² + a y - (b + 2).

We need f(y) = 0 to have at least one root in (-∞, -2] ∪ [2, ∞).

To find this, we can analyze the quadratic function f(y). Since it's a quadratic opening upwards (coefficient of y² is 1 > 0), the graph is a parabola opening upwards. Therefore, the equation f(y) = 0 will have roots in (-∞, -2] ∪ [2, ∞) if either:

1. The smaller root is ≤ -2, or

2. The larger root is ≥ 2.

But since the parabola opens upwards, if the smaller root is ≤ -2, then the entire left side of the parabola will be negative up to that root, but since it's opening upwards, the function tends to +infty as y approaches ±infty. Wait, maybe I need to think more carefully.

Alternatively, for the quadratic equation to have at least one root in (-∞, -2] ∪ [2, ∞), we can use the following approach:

Either:

- The quadratic has a root ≥ 2, or

- The quadratic has a root ≤ -2.

But since the quadratic is opening upwards, if there is a root ≥ 2, then f(2) ≤ 0, because the parabola crosses the y-axis at f(2) and since it opens upwards, if there is a root ≥ 2, then f(2) ≤ 0. Similarly, if there is a root ≤ -2, then f(-2) ≤ 0.

Wait, let me recall the concept here. For a quadratic equation f(y) = 0 with leading coefficient positive, if there is a root ≥ k, then f(k) ≤ 0. Similarly, if there is a root ≤ m, then f(m) ≤ 0. Because if the parabola crosses the x-axis at some point ≥ k, then at y = k, the function is below or equal to zero. Similarly for the other side.

Therefore, the condition that the quadratic has at least one root in (-∞, -2] ∪ [2, ∞) is equivalent to f(-2) ≤ 0 or f(2) ≤ 0. Wait, but is that sufficient?

Wait, suppose f(-2) ≤ 0. Then since the parabola opens upwards, there must be a root ≤ -2? Wait, not necessarily. Because if f(-2) ≤ 0, then since the parabola opens upwards, there is a root to the left of -2 or between -2 and the other root. Wait, maybe I need to think more carefully.

Let me recall that for a quadratic equation f(y) = 0 with leading coefficient positive, the following holds:

- If f(k) ≤ 0, then there is at least one root in (-∞, k] if the parabola crosses the x-axis to the left of k. Wait, no. Actually, if f(k) ≤ 0, then since the parabola opens upwards, there must be at least one root ≤ k and another root ≥ k? No, that's not necessarily true. For example, if the entire parabola is below zero at y = k, but since it opens upwards, it must cross the x-axis on both sides of k if there are two real roots. Wait, but if f(k) ≤ 0, then there must be at least one root ≤ k and another root ≥ k? Wait, no. Suppose the quadratic has two real roots. If f(k) ≤ 0, then k lies between the two roots. Because if the parabola opens upwards, then between the two roots, the function is negative. So if f(k) ≤ 0, then k is between the two roots. Therefore, if f(-2) ≤ 0, then -2 is between the two roots, meaning that one root is ≤ -2 and the other is ≥ -2. Similarly, if f(2) ≤ 0, then 2 is between the two roots, meaning one root is ≤ 2 and the other is ≥ 2.

But we need at least one root in (-∞, -2] ∪ [2, ∞). So, if f(-2) ≤ 0, then there is a root ≤ -2 (since one root is ≤ -2 and the other is ≥ -2). Similarly, if f(2) ≤ 0, then there is a root ≥ 2. Therefore, the condition that the quadratic has at least one root in (-∞, -2] ∪ [2, ∞) is equivalent to f(-2) ≤ 0 or f(2) ≤ 0. Wait, but is that correct?

Wait, suppose f(-2) ≤ 0. Then, since the parabola opens upwards, there are two real roots, one ≤ -2 and one ≥ -2. Therefore, there is a root ≤ -2. Similarly, if f(2) ≤ 0, then there are two real roots, one ≤ 2 and one ≥ 2, so there is a root ≥ 2. Therefore, the condition that the quadratic has at least one root in (-∞, -2] ∪ [2, ∞) is equivalent to f(-2) ≤ 0 or f(2) ≤ 0. However, we also need to check if there are cases where both f(-2) > 0 and f(2) > 0, but the quadratic still has a root in (-∞, -2] ∪ [2, ∞). Is that possible?

Wait, suppose that the quadratic has two real roots, both greater than 2. Then f(2) would be positive, because the parabola is opening upwards, so at y=2, which is to the left of both roots, the function is positive. Similarly, if both roots are less than -2, then f(-2) would be positive. Therefore, if the quadratic has both roots in (-∞, -2] ∪ [2, ∞), then f(-2) and f(2) would be positive? Wait, no. If both roots are greater than 2, then f(2) is positive. If both roots are less than -2, then f(-2) is positive. However, if one root is less than -2 and the other is greater than 2, then f(-2) and f(2) would be negative? Wait, let me think.

Suppose the quadratic has one root less than -2 and one root greater than 2. Then, since the parabola opens upwards, at y = -2, which is between the two roots, the function f(-2) would be negative. Similarly, at y = 2, which is between the two roots, f(2) would be negative. Therefore, in this case, both f(-2) and f(2) are negative. Therefore, if the quadratic has one root in (-∞, -2] and one root in [2, ∞), then both f(-2) ≤ 0 and f(2) ≤ 0. However, if the quadratic has both roots in (-∞, -2], then f(-2) would be positive (since the parabola is above zero at y=-2, as both roots are to the left of -2). Similarly, if both roots are in [2, ∞), then f(2) is positive. Therefore, the condition that the quadratic has at least one root in (-∞, -2] ∪ [2, ∞) is equivalent to f(-2) ≤ 0 or f(2) ≤ 0. Because if f(-2) ≤ 0, then there is a root ≤ -2, and if f(2) ≤ 0, there is a root ≥ 2. However, if both f(-2) ≤ 0 and f(2) ≤ 0, then there are roots in both intervals. Therefore, the set of (a, b) such that the original equation has a real root is the set of (a, b) where f(-2) ≤ 0 or f(2) ≤ 0.

Therefore, we can write:

Either f(-2) ≤ 0 or f(2) ≤ 0.

So let's compute f(-2) and f(2):

First, f(y) = y² + a y - (b + 2).

Therefore:

f(-2) = (-2)² + a*(-2) - (b + 2) = 4 - 2a - b - 2 = 2 - 2a - b.

Similarly, f(2) = (2)² + a*(2) - (b + 2) = 4 + 2a - b - 2 = 2 + 2a - b.

Therefore, the conditions are:

Either 2 - 2a - b ≤ 0 or 2 + 2a - b ≤ 0.

So, simplifying:

First condition: 2 - 2a - b ≤ 0 ⇒ 2a + b ≥ 2.

Second condition: 2 + 2a - b ≤ 0 ⇒ 2a - b ≤ -2 ⇒ b ≥ 2a + 2.

But since a and b are between 0 and 1, let's analyze these inequalities.

First, for the first condition: 2a + b ≥ 2.

Given that a and b are in [0,1], let's see what region this defines.

Similarly, for the second condition: b ≥ 2a + 2. However, since b ≤ 1, and 2a + 2 ≥ 2*0 + 2 = 2, but b ≤ 1 < 2, so this inequality b ≥ 2a + 2 can never be satisfied. Therefore, the second condition is impossible. Therefore, the only condition we need is 2a + b ≥ 2.

Wait, that seems important. Let me check again.

Second condition: 2 + 2a - b ≤ 0 ⇒ 2a - b ≤ -2 ⇒ b ≥ 2a + 2. Since a is between 0 and 1, 2a is between 0 and 2, so 2a + 2 is between 2 and 4. However, since b is between 0 and 1, b cannot be ≥ 2a + 2, because 2a + 2 ≥ 2 > 1. Therefore, the second condition is impossible. Therefore, the only condition is 2a + b ≥ 2.

Therefore, the set S is the set of (a, b) in [0,1]x[0,1] such that 2a + b ≥ 2.

Wait, but hold on. Let me confirm this conclusion. Because if the second condition is impossible, then the only way for the quadratic to have a root in (-∞, -2] ∪ [2, ∞) is if 2a + b ≥ 2. Therefore, the set S is the region in the unit square where 2a + b ≥ 2. Therefore, the area of S is the area of the region defined by 2a + b ≥ 2 with 0 ≤ a, b ≤ 1.

But let me check if this is correct. Let me think again. Because earlier, I thought that if f(-2) ≤ 0 or f(2) ≤ 0, but since f(2) ≤ 0 is impossible, only f(-2) ≤ 0 is possible. Therefore, the condition reduces to 2 - 2a - b ≤ 0 ⇒ 2a + b ≥ 2. Therefore, the set S is the set of (a, b) with 2a + b ≥ 2, 0 ≤ a, b ≤ 1.

But let me check with an example. Let me take a = 1, b = 0. Then 2a + b = 2*1 + 0 = 2, which satisfies the condition. Let me check if the original equation has a real root. The equation becomes x⁴ + x³ - 0x² + x + 1 = x⁴ + x³ + x + 1. Let me try to see if this has real roots. Let me plug in x = -1: (-1)^4 + (-1)^3 + (-1) + 1 = 1 - 1 - 1 + 1 = 0. So x = -1 is a root. Therefore, yes, there is a real root. So that point (1, 0) is in S.

Another example: take a = 0.5, b = 1. Then 2a + b = 1 + 1 = 2, which satisfies the condition. Let me check the equation: x⁴ + 0.5x³ - 1x² + 0.5x + 1. Let me try x = -1: 1 - 0.5 - 1 - 0.5 + 1 = 0. So again, x = -1 is a root. So that works.

Another example: take a = 0. Let me see. If a = 0, then 2a + b = b. So to satisfy 2a + b ≥ 2, we need b ≥ 2. But since b ≤ 1, there are no solutions here. Therefore, when a = 0, there are no points in S. Which makes sense, because if a = 0, the equation becomes x⁴ - b x² + 1 = 0. Let me check if this has real roots. Let me set y = x², then equation becomes y² - b y + 1 = 0. The discriminant is b² - 4. Since b ≤ 1, discriminant is ≤ 1 - 4 = -3 < 0. Therefore, no real roots. Therefore, when a = 0, there are no real roots, which matches our previous conclusion.

Another example: take a = 1, b = 1. Then 2a + b = 2 + 1 = 3 ≥ 2. Let me check the equation: x⁴ + x³ - x² + x + 1. Let me try x = -1: 1 - 1 - 1 - 1 + 1 = -1 ≠ 0. Hmm, so x = -1 is not a root here. Wait, but according to our previous logic, since 2a + b ≥ 2, there should be a real root. Maybe I need to check another value. Let me try x = -2: (-2)^4 + (-2)^3 - (-2)^2 + (-2) + 1 = 16 - 8 - 4 - 2 + 1 = 3 ≠ 0. Hmm. Maybe there's another real root?

Alternatively, maybe I need to check the transformed equation. Let me use the substitution y = x + 1/x. For the equation x⁴ + x³ - x² + x + 1 = 0. Dividing by x²: x² + x - 1 + 1/x + 1/x² = 0. Which is y² - 2 + a y - b = 0? Wait, earlier we had y² + a y - (b + 2) = 0. Wait, in this case, a = 1, b = 1. Therefore, the equation is y² + y - (1 + 2) = y² + y - 3 = 0. Solving this quadratic: y = [-1 ± √(1 + 12)] / 2 = [-1 ± √13]/2. So the roots are approximately [-1 + 3.605]/2 ≈ 1.3025 and [-1 - 3.605]/2 ≈ -2.3025. Now, since y must be in (-∞, -2] ∪ [2, ∞). The first root is ≈1.3025 which is between -2 and 2, so not in the required range. The second root is ≈-2.3025 which is less than -2. Therefore, there is a real root corresponding to y ≈ -2.3025. Therefore, there exists real x such that x + 1/x ≈ -2.3025. Let me check if such x exists. For example, solving x + 1/x = -2.3025. Let me set z = x, then z + 1/z = -k where k > 0. Multiply both sides by z: z² + 1 = -k z ⇒ z² + k z + 1 = 0. The discriminant is k² - 4. For real solutions, we need k² - 4 ≥ 0 ⇒ k ≥ 2. In our case, k ≈ 2.3025, which is greater than 2, so there are real solutions. Therefore, there are real roots. Therefore, even though x = -1 is not a root, there are other real roots. Therefore, the point (1, 1) is indeed in S. Therefore, our condition seems correct.

Therefore, the set S is defined by 2a + b ≥ 2 with 0 ≤ a, b ≤ 1. Therefore, the area of S is the area of the region in the unit square where 2a + b ≥ 2.

Now, let me visualize this region. The line 2a + b = 2 in the a-b plane (with a and b between 0 and 1). Let me find the intercepts.

When a = 0: 0 + b = 2 ⇒ b = 2, but since b ≤ 1, this is outside the square.

When b = 0: 2a + 0 = 2 ⇒ a = 1. So the line passes through (1, 0).

When a = 1: 2*1 + b = 2 ⇒ b = 0. So the line passes through (1, 0) and (0, 2), but since b can't be 2, the line within the square goes from (1, 0) to where? Let me find another point. Let me solve for when b = 1: 2a + 1 = 2 ⇒ 2a = 1 ⇒ a = 0.5. Therefore, the line passes through (0.5, 1). Therefore, within the unit square, the line 2a + b = 2 connects (1, 0) to (0.5, 1). Therefore, the region 2a + b ≥ 2 is the area above this line within the square.

Therefore, the area of S is the area above the line from (1, 0) to (0.5, 1) in the unit square.

To compute this area, we can compute the area of the region in the unit square where 2a + b ≥ 2. This is a triangle or a trapezoid?

Let me sketch mentally. The line goes from (1, 0) to (0.5, 1). The region 2a + b ≥ 2 is above this line. So, in the unit square, the area above the line is a triangle with vertices at (1, 0), (0.5, 1), and (0, 1)? Wait, no. Wait, when a = 0, b must be ≥ 2, which is outside the square. So the region within the square is bounded by the line from (1, 0) to (0.5, 1), and then the top edge of the square from (0.5, 1) to (0, 1), and the left edge from (0, 1) to (0, 0)? Wait, no. Wait, actually, the region 2a + b ≥ 2 in the square is a polygon with vertices at (1, 0), (0.5, 1), and (0, 1)? Wait, let me check.

At b = 1, the line 2a + 1 = 2 ⇒ a = 0.5. So the line intersects the top edge of the square (b=1) at (0.5, 1). Then, if we go to the left, when a decreases, b would have to increase to satisfy 2a + b ≥ 2, but since b can't exceed 1, the region is bounded by the line from (1, 0) to (0.5, 1), and then the line from (0.5, 1) to (0, 1) along b=1? Wait, but for a < 0.5, even if b=1, 2a + 1 ≥ 2 ⇒ 2a ≥ 1 ⇒ a ≥ 0.5. Therefore, for a < 0.5, even with b=1, 2a + b < 2. Therefore, the region 2a + b ≥ 2 is only above the line from (1, 0) to (0.5, 1), and for a between 0.5 and 1, b between 2 - 2a and 1. Therefore, the area is a triangle with vertices at (1, 0), (0.5, 1), and (1, 1). Wait, let me check.

Wait, when a = 1, b can be from 0 to 1, but 2a + b ≥ 2 ⇒ 2*1 + b ≥ 2 ⇒ b ≥ 0. So for a = 1, all b between 0 and 1 satisfy the condition. Wait, but earlier when we checked a = 1, b = 0, it worked. But when a = 1, b can be from 0 to 1. However, according to the line 2a + b = 2, when a = 1, b = 0. So the region for a = 1 is from b = 0 to 1. Similarly, for a between 0.5 and 1, the lower bound for b is 2 - 2a, and upper bound is 1. For a between 0 and 0.5, there is no b in [0,1] that satisfies 2a + b ≥ 2, since 2a ≤ 1, so 2a + b ≤ 1 + 1 = 2, but equality only when a = 0.5 and b = 1. Therefore, the region is a trapezoid? Wait, no. Wait, for a from 0.5 to 1, the region is between b = 2 - 2a and b = 1. So the area can be calculated as the integral from a = 0.5 to 1 of [1 - (2 - 2a)] da. Which is integral from 0.5 to 1 of (2a - 1) da.

Alternatively, since it's a triangle. Let me think. The region is a triangle with vertices at (1, 0), (1, 1), and (0.5, 1). Because when a = 1, b ranges from 0 to 1, and when a = 0.5, b ranges from 1 to 1 (i.e., just the point (0.5, 1)). Therefore, the area is the area of the triangle with vertices at (1, 0), (1, 1), and (0.5, 1). The base of this triangle is from (1, 0) to (1, 1), which is length 1. The height is from (1, 1) to (0.5, 1), which is 0.5. But since it's a right triangle? Wait, no. Wait, the three points are (1, 0), (1, 1), and (0.5, 1). So plotting these points, it's a right triangle with legs of length 1 (from (1,0) to (1,1)) and 0.5 (from (1,1) to (0.5,1)). Therefore, the area is (base * height)/2 = (1 * 0.5)/2 = 0.25. Wait, but let me check with integration.

Compute the area as integral from a = 0.5 to 1 of [upper bound - lower bound] da. Upper bound is 1, lower bound is 2 - 2a. Therefore, the integral is ∫_{0.5}^1 [1 - (2 - 2a)] da = ∫_{0.5}^1 (2a - 1) da.

Compute the integral:

∫ (2a - 1) da = a² - a evaluated from 0.5 to 1.

At a = 1: 1² - 1 = 0.

At a = 0.5: (0.5)² - 0.5 = 0.25 - 0.5 = -0.25.

Therefore, the integral is 0 - (-0.25) = 0.25. Therefore, the area is 0.25. Therefore, the area of S is 1/4.

Wait, but let me think again. If the region is a triangle with vertices at (1, 0), (1, 1), and (0.5, 1), then the area is indeed 1/2 * base * height. The base can be considered as the vertical segment from (1, 0) to (1, 1), which is length 1. The height is the horizontal distance from this base to the point (0.5, 1), which is 0.5. However, since it's a right triangle, the area is (base * height)/2 = (1 * 0.5)/2 = 0.25. So that's 1/4. Therefore, the area is 1/4.

But wait, let me check with another approach. The line 2a + b = 2 cuts the square. The area above the line is a triangle with area 1/4. Therefore, the answer is 1/4? But let me check with another example. Suppose we have the unit square [0,1]x[0,1]. The line 2a + b = 2 goes from (1, 0) to (0.5, 1). The area above this line is the triangle we discussed. Alternatively, the area below the line is the rest of the square. The area below the line would be the area of the trapezoid from (0,0) to (1,0) to (0.5,1) to (0,1). Wait, but maybe not. Let me compute the area below the line 2a + b = 2 in the square. For a from 0 to 0.5, b can go up to 1, but the line 2a + b = 2 would have b = 2 - 2a. However, for a from 0 to 0.5, 2 - 2a ranges from 2 to 1. Therefore, the area below the line is the area under the line from a=0 to a=0.5, plus the area from a=0.5 to a=1, where the line is above b=1. Wait, this is getting complicated. Alternatively, since the total area of the square is 1, and the area above the line is 1/4, then the area below the line would be 3/4. But let me check with integrating.

Alternatively, the area above the line 2a + b ≥ 2 is the integral from a=0.5 to 1 of (1 - (2 - 2a)) da = 0.25 as before. Therefore, the area is 1/4. Therefore, the answer is 1/4? But let me check with another method.

Alternatively, think of the region 2a + b ≥ 2 in the unit square. This is equivalent to b ≥ 2 - 2a. So for each a between 0.5 and 1, b ranges from 2 - 2a to 1. For a between 0 and 0.5, there is no b in [0,1] that satisfies b ≥ 2 - 2a, since 2 - 2a ≥ 1 when a ≤ 0.5. Therefore, the area is the integral from a=0.5 to 1 of [1 - (2 - 2a)] da = integral from 0.5 to 1 of (2a - 1) da = [a² - a] from 0.5 to 1 = (1 - 1) - (0.25 - 0.5) = 0 - (-0.25) = 0.25. Therefore, 0.25 is 1/4. Therefore, the area is 1/4.

But wait, the problem says "the set of points (a,b) with 0 ≤ a, b ≤ 1 such that the equation... has at least one real root." So according to our analysis, this area is 1/4. However, let me check again if there are other cases where the equation has real roots. Earlier, we considered only the case where the quadratic in y has a root in (-∞, -2] ∪ [2, ∞). But is there a possibility that even if f(-2) > 0 and f(2) > 0, the quadratic could still have a root in (-∞, -2] ∪ [2, ∞)? For example, if the quadratic has two real roots, both outside of (-2, 2). Wait, but if the quadratic has two real roots, then since the parabola opens upwards, if both roots are greater than 2, then f(2) > 0. Similarly, if both roots are less than -2, then f(-2) > 0. Therefore, if the quadratic has two real roots outside of (-2, 2), then both f(-2) and f(2) are positive. Therefore, in that case, the equation would have real roots, but our condition f(-2) ≤ 0 or f(2) ≤ 0 would not be satisfied. Therefore, our previous conclusion that the condition is equivalent to f(-2) ≤ 0 or f(2) ≤ 0 is incomplete? Wait, but earlier we thought that if the quadratic has a root in (-∞, -2] ∪ [2, ∞), then either f(-2) ≤ 0 or f(2) ≤ 0. But if the quadratic has two roots outside of (-2, 2), then both f(-2) and f(2) are positive. Therefore, our previous conclusion was wrong? Wait, this seems contradictory. Let me think again.

Let me take an example where the quadratic has two roots outside of (-2, 2). For example, suppose the quadratic equation y² + a y - (b + 2) = 0 has two roots both greater than 2. Then, since the parabola opens upwards, f(2) = 2 + 2a - b. If both roots are greater than 2, then f(2) must be positive. Similarly, if both roots are less than -2, then f(-2) is positive. Therefore, in such cases, the quadratic does have real roots outside of (-2, 2), but our condition f(-2) ≤ 0 or f(2) ≤ 0 is not satisfied. Therefore, our previous conclusion was wrong. Therefore, there is a mistake in the reasoning.

Wait, so where is the mistake? Earlier, I thought that if the quadratic has a root in (-∞, -2] ∪ [2, ∞), then f(-2) ≤ 0 or f(2) ≤ 0. But actually, if the quadratic has two roots outside of (-2, 2), then both f(-2) and f(2) are positive. Therefore, the condition that the quadratic has at least one root in (-∞, -2] ∪ [2, ∞) is not equivalent to f(-2) ≤ 0 or f(2) ≤ 0. Therefore, my previous conclusion was wrong. Therefore, I need to re-examine the problem.

So, going back. The quadratic equation y² + a y - (b + 2) = 0 has real roots if discriminant is non-negative, which it always is. The roots are y = [-a ± √(a² + 4(b + 2))]/2. We need at least one of these roots to be in (-∞, -2] ∪ [2, ∞). Therefore, we need to find conditions on a and b such that at least one of the roots is in that interval.

To find this, we can use the following approach: For a quadratic equation f(y) = 0, the condition that at least one root is ≥ 2 is equivalent to f(2) ≤ 0 or the quadratic has a root ≥ 2. Similarly for ≤ -2. However, since the quadratic opens upwards, the condition that there is a root ≥ 2 is equivalent to f(2) ≤ 0 or the vertex is to the right of 2 and the minimum value is ≤ 0. Wait, this is getting complicated. Maybe another approach is better.

Alternatively, for the quadratic equation y² + a y - (b + 2) = 0, let me denote the roots as y₁ and y₂. We need at least one of y₁ or y₂ to be ≤ -2 or ≥ 2.

This can be approached by considering the following cases:

Case 1: At least one root ≥ 2.

Case 2: At least one root ≤ -2.

But these cases can overlap. However, since the quadratic is symmetric in some way, maybe we can find conditions for each case.

For Case 1: At least one root ≥ 2.

This occurs if either:

- The quadratic has a root exactly at 2, or

- The quadratic has a root greater than 2.

For a quadratic with leading coefficient positive, this happens if f(2) ≤ 0 or the vertex is to the right of 2 and the minimum value is ≤ 0. Wait, but since the parabola opens upwards, if f(2) ≤ 0, then there is a root ≥ 2. If f(2) > 0, then both roots are either less than 2 or one less and one greater. Wait, no. If f(2) > 0, then since the parabola opens upwards, if there are two real roots, they are both on the same side of 2? No, if f(2) > 0, then depending on the position of the vertex, the roots could be both less than 2 or both greater than 2? Wait, no. If the parabola opens upwards and f(2) > 0, then if there are two real roots, they are either both less than 2 or both greater than 2? No, that's not necessarily true. For example, take f(y) = y² - 4. Then f(2) = 0, which is the case of a root at 2. If f(y) = y² - 3, then f(2) = 4 - 3 = 1 > 0, and the roots are ±√3 ≈ ±1.732, which are both less than 2. If f(y) = y² - 5, then f(2) = 4 - 5 = -1 < 0, so there are roots on both sides of 2. Wait, no. Wait, if f(2) < 0, then since the parabola opens upwards, there are two real roots, one less than 2 and one greater than 2. If f(2) = 0, then one root is at 2. If f(2) > 0, then either both roots are less than 2 or both are greater than 2? Wait, no. If f(2) > 0, then depending on the vertex, it could be that both roots are less than 2 or both are greater than 2. For example, take f(y) = y² - 3, which has roots at ±√3 ≈ ±1.732. So both roots are less than 2. If we take f(y) = y² - 5, which has roots at ±√5 ≈ ±2.236. So one root is less than 2 and one is greater than 2. Wait, no, √5 ≈ 2.236 > 2, and -√5 ≈ -2.236 < -2. So in this case, one root is less than -2, and one is greater than 2. Wait, but f(2) = 4 - 5 = -1 < 0. So if f(2) < 0, then there is one root less than 2 and one greater than 2. If f(2) > 0, then both roots are either less than 2 or both greater than 2? Wait, no. For example, take f(y) = y² - 4y + 3. This factors as (y - 1)(y - 3). So roots at 1 and 3. f(2) = 4 - 8 + 3 = -1 < 0. So here, f(2) < 0, and there are roots on both sides of 2. If I take f(y) = y² - 4y + 5. Then discriminant is 16 - 20 = -4 < 0, no real roots. If I take f(y) = y² - 4y + 4 = (y - 2)^2. Then f(2) = 0, and there's a double root at 2. If I take f(y) = y² - 4y + 6. Discriminant is 16 - 24 = -8 < 0. No real roots. If I take f(y) = y² - 4y + 2. Discriminant is 16 - 8 = 8 > 0. Roots at [4 ± √8]/2 = 2 ± √2 ≈ 2 ± 1.414. So one root ≈ 3.414 > 2, and one root ≈ 0.586 < 2. Therefore, f(2) = 4 - 8 + 2 = -2 < 0. So again, when f(2) < 0, there are roots on both sides of 2. If f(2) > 0, then there are no real roots or both roots on one side. Wait, but if f(2) > 0, and there are real roots, then they must be both on one side of 2. For example, take f(y) = y² - 3, which has roots at ±√3 ≈ ±1.732. Both less than 2. If I take f(y) = y² - 5, which has roots at ±√5 ≈ ±2.236. So one root is less than -2, and one is greater than 2. Wait, but f(2) = 4 - 5 = -1 < 0. So in this case, f(2) < 0. Therefore, if f(2) > 0, then there are no real roots or both roots are on the same side of 2. Wait, but how can there be real roots with f(2) > 0 and both roots on the same side of 2? For example, take f(y) = y² - 4y + 5. Wait, discriminant is negative. How about f(y) = y² - 4y + 6. Discriminant negative. How about f(y) = y² - 4y + 4. Then f(2) = 0, double root at 2. If I take f(y) = y² - 4y + 3. Then f(2) = -1 < 0. So maybe if f(2) > 0, then there are no real roots or both roots are on the same side of 2. Wait, but how can there be real roots with f(2) > 0 and both roots on the same side of 2? For example, take f(y) = y² - 3. Then f(2) = 4 - 3 = 1 > 0, and roots at ±√3 ≈ ±1.732. Both roots are less than 2. So yes, in this case, both roots are less than 2. Similarly, if I take f(y) = y² - 5, then f(2) = 4 - 5 = -1 < 0, and roots at ±√5 ≈ ±2.236. So one root is less than -2, one is greater than 2. Therefore, if f(2) > 0, then both roots are less than 2 or both are greater than 2? Wait, no. If f(2) > 0, and there are real roots, then they are both less than 2 or both greater than 2? How can they be both greater than 2? For example, take f(y) = y² - 4y + 5. Wait, discriminant is negative. How about f(y) = y² - 4y + 6. Still discriminant negative. How about f(y) = y² - 4y + 7. Still discriminant negative. Wait, maybe it's not possible to have both roots greater than 2 with f(2) > 0? Let me think. Suppose we have a quadratic with both roots greater than 2. Then, since the parabola opens upwards, f(2) would be positive, because at y = 2, which is to the left of both roots, the function is positive. Therefore, yes, if both roots are greater than 2, then f(2) > 0. Similarly, if both roots are less than 2, then f(2) > 0. Therefore, the condition that there is at least one root ≥ 2 is equivalent to f(2) ≤ 0 or the quadratic has both roots ≥ 2. Wait, but how do we check for both roots ≥ 2?

For a quadratic to have both roots ≥ 2, the following conditions must be satisfied:

1. The quadratic has real roots (discriminant ≥ 0).

2. The axis of symmetry is ≥ 2.

3. f(2) ≥ 0.

Wait, the axis of symmetry is at y = -a/2. For both roots to be ≥ 2, the axis of symmetry must be ≥ 2, and f(2) ≥ 0. Wait, no. Let me recall the conditions for both roots ≥ k:

For a quadratic f(y) = y² + py + q, to have both roots ≥ k, the following must hold:

1. Discriminant ≥ 0.

2. The axis of symmetry ≥ k.

3. f(k) ≥ 0.

Similarly, for both roots ≤ k, the axis of symmetry ≤ k and f(k) ≥ 0.

Therefore, applying this to our case, for both roots ≥ 2:

1. Discriminant ≥ 0 (which it always is).

2. Axis of symmetry ≥ 2. The axis of symmetry is at y = -a/2. So -a/2 ≥ 2 ⇒ -a ≥ 4 ⇒ a ≤ -4. But since a is between 0 and 1, this is impossible. Therefore, there are no such quadratics with both roots ≥ 2. Therefore, the only way to have at least one root ≥ 2 is if f(2) ≤ 0. Similarly, for at least one root ≤ -2, we need f(-2) ≤ 0. Therefore, our original conclusion was correct! Because for the quadratic to have at least one root ≥ 2, since the axis of symmetry cannot be ≥ 2 (as a ≥ 0), the only possibility is f(2) ≤ 0. Similarly, for at least one root ≤ -2, since the axis of symmetry is at y = -a/2, which for a ≥ 0 is ≤ 0, so to have a root ≤ -2, we need f(-2) ≤ 0. Therefore, the condition that the quadratic has at least one root in (-∞, -2] ∪ [2, ∞) is equivalent to f(-2) ≤ 0 or f(2) ≤ 0. And since f(2) ≤ 0 is equivalent to 2 + 2a - b ≤ 0 ⇒ b ≥ 2a + 2, which is impossible, and f(-2) ≤ 0 is equivalent to 2 - 2a - b ≤ 0 ⇒ 2a + b ≥ 2. Therefore, the only condition is 2a + b ≥ 2. Therefore, the area is indeed 1/4.

But earlier, I thought there was a contradiction, but now it seems that the previous conclusion was correct. The confusion was due to thinking that there could be cases where both roots are outside of (-2, 2), but in reality, for our specific quadratic, since the axis of symmetry is at y = -a/2, which for a ≥ 0 is ≤ 0, so the quadratic cannot have both roots ≥ 2. Therefore, the only way to have a root ≥ 2 is f(2) ≤ 0, which is impossible, and the only way to have a root ≤ -2 is f(-2) ≤ 0. Therefore, the condition is indeed 2a + b ≥ 2, and the area is 1/4.

Therefore, the answer should be 1/4, so the area of S is 1/4, which is 0.25. Therefore, the final answer is \boxed{\dfrac{1}{4}}.

**Final Answer**
\boxed{\dfrac{1}{4}}

To determine the area of the set $ S $, we are given the condition that the quartic polynomial

$$
x^4 + ax^3 - bx^2 + ax + 1 = 0
$$

has at least one real root, for $ (a, b) \in [0,1] \times [0,1] $. Our goal is to find the area of the region in the unit square where this condition is satisfied.

---

### Step 1: Recognize the Polynomial Structure

The polynomial is symmetric in a special way. It is **palindromic**, meaning its coefficients read the same forwards and backwards. This allows us to use a substitution to simplify the equation.

Divide the equation by $ x^2 $ (valid for $ x \ne 0 $):

$$
x^2 + a x - b + \frac{a}{x} + \frac{1}{x^2} = 0
$$

Let $ y = x + \frac{1}{x} $. Then, we know:

$$
x^2 + \frac{1}{x^2} = y^2 - 2
$$

Substituting into the equation:

$$
y^2 - 2 + a y - b = 0 \quad \Rightarrow \quad y^2 + a y - (b + 2) = 0
$$

This is a quadratic in $ y $, and we are interested in whether this equation has a real solution $ y $ such that $ y = x + \frac{1}{x} $ for some real $ x \ne 0 $.

---

### Step 2: Determine the Range of $ y = x + \frac{1}{x} $

For real $ x \ne 0 $, the expression $ x + \frac{1}{x} $ has the range:

$$
(-\infty, -2] \cup [2, \infty)
$$

This is because:

- For $ x > 0 $, by AM-GM, $ x + \frac{1}{x} \ge 2 $
- For $ x < 0 $, $ x + \frac{1}{x} \le -2 $

So, the original equation has a real root if and only if the quadratic in $ y $ has at least one root in $ (-\infty, -2] \cup [2, \infty) $.

---

### Step 3: Analyze the Quadratic in $ y $

Let the quadratic be:

$$
f(y) = y^2 + a y - (b + 2)
$$

We want $ f(y) = 0 $ to have at least one root in $ (-\infty, -2] \cup [2, \infty) $.

Since the parabola opens upwards (coefficient of $ y^2 $ is positive), the condition for having a root in $ (-\infty, -2] \cup [2, \infty) $ is equivalent to:

- $ f(-2) \le 0 $ or $ f(2) \le 0 $

Let’s compute these:

- $ f(-2) = (-2)^2 + a(-2) - (b + 2) = 4 - 2a - b - 2 = 2 - 2a - b $
- $ f(2) = 2^2 + a(2) - (b + 2) = 4 + 2a - b - 2 = 2 + 2a - b $

So the conditions become:

- $ 2 - 2a - b \le 0 \Rightarrow 2a + b \ge 2 $
- $ 2 + 2a - b \le 0 \Rightarrow b \ge 2a + 2 $

Now, since $ a, b \in [0,1] $, the second inequality $ b \ge 2a + 2 $ is **never satisfied**, because the right-hand side is at least 2, while $ b \le 1 $. So the only valid condition is:

$$
2a + b \ge 2
$$

---

### Step 4: Determine the Region in the Unit Square

We now find the area of the region in the unit square $ [0,1] \times [0,1] $ where $ 2a + b \ge 2 $.

This is a linear inequality. The line $ 2a + b = 2 $ intersects the square at:

- When $ a = 0 $: $ b = 2 $ → outside the square
- When $ b = 0 $: $ a = 1 $
- When $ b = 1 $: $ 2a + 1 = 2 \Rightarrow a = 0.5 $

So the line passes through the points $ (1, 0) $ and $ (0.5, 1) $, and the region $ 2a + b \ge 2 $ lies **above** this line within the square.

This region is a **right triangle** with vertices at:

- $ (1, 0) $
- $ (1, 1) $
- $ (0.5, 1) $

The base of the triangle is from $ (1, 0) $ to $ (1, 1) $, which has length 1.

The height is from $ (1, 1) $ to $ (0.5, 1) $, which has length $ 0.5 $.

So the area is:

$$
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 0.5 = \frac{1}{4}
$$

---

### Final Answer

$$
\boxed{\dfrac{1}{4}}
$$

[Assistant]
Explanation:
Okay, so I need to find the area of the set S, which consists of points (a, b) where 0 ≤ a, b ≤ 1, such that the equation x⁴ + a x³ - b x² + a x + 1 = 0 has at least one real root. Hmm, let me think about how to approach this.

First, the equation is a quartic polynomial. Quartic equations can be tricky, but maybe there's some symmetry or substitution that can simplify things. Let me write down the equation again:

x⁴ + a x³ - b x² + a x + 1 = 0.

Looking at the coefficients, I notice that the coefficients of x⁴ and the constant term are both 1, and the coefficients of x³ and x are both a. The middle term is -b x². Maybe there's some symmetry here. Let me check if the polynomial is palindromic or something like that.

A palindromic polynomial has coefficients that read the same forwards and backwards. Let me check:

Original coefficients: 1 (x⁴), a (x³), -b (x²), a (x), 1 (constant). So yes! It's a palindromic polynomial. That means if we reverse the coefficients, we get the same polynomial. For palindromic polynomials of even degree, there's a substitution that can be used. Specifically, for a palindromic polynomial of degree 4, we can divide both sides by x² to make it symmetric.

Let me try that. Dividing the equation by x² (assuming x ≠ 0, but we can check x=0 separately later):

x² + a x - b + a / x + 1 / x² = 0.

Now, let me set y = x + 1/x. Then, note that y² = x² + 2 + 1/x², so x² + 1/x² = y² - 2. Similarly, x³ + 1/x³ = y³ - 3y, but maybe I don't need that here.

Looking at the equation:

x² + 1/x² + a(x + 1/x) - b = 0.

Which is:

(y² - 2) + a y - b = 0.

So, simplifying:

y² + a y - (b + 2) = 0.

Therefore, the original equation can be transformed into a quadratic in terms of y. However, this substitution is valid only when x ≠ 0. But since we are looking for real roots, we need to check if x=0 is a root. Let me check x=0:

Plugging x=0 into the original equation: 0 + 0 - 0 + 0 + 1 = 1 ≠ 0. So x=0 is not a root. Therefore, all real roots satisfy x ≠ 0, so the substitution is valid for all real roots.

Therefore, the original equation has a real root if and only if the equation y² + a y - (b + 2) = 0 has a real solution y such that y = x + 1/x for some real x ≠ 0.

But we need to relate this back to real x. So, for each real solution y of the quadratic equation, we need to check if there exists a real x ≠ 0 such that y = x + 1/x.

So, first, let's analyze the quadratic equation in y:

y² + a y - (b + 2) = 0.

Let me denote this as:

y² + a y - (b + 2) = 0.

The solutions for y are:

y = [-a ± √(a² + 4(b + 2))]/2.

But for real solutions y, the discriminant must be non-negative. However, even if the discriminant is non-negative, we need to check if the corresponding y can be expressed as x + 1/x for some real x.

So, first, the discriminant D = a² + 4(b + 2) must be ≥ 0. But since a and b are between 0 and 1, let's check:

a² is between 0 and 1, and 4(b + 2) is between 4(0 + 2) = 8 and 4(1 + 2) = 12. Therefore, D is always positive. So the quadratic equation in y always has two real roots. Therefore, the original equation will have real roots if and only if at least one of these y's can be expressed as x + 1/x for some real x ≠ 0.

So, now, we need to find for which (a, b) there exists a real number y such that y = x + 1/x and y satisfies the quadratic equation above.

But what is the range of possible y = x + 1/x for real x ≠ 0?

Let me recall that for real x ≠ 0, the expression x + 1/x has a range of (-∞, -2] ∪ [2, ∞). This is because, for x > 0, by AM ≥ GM, x + 1/x ≥ 2, and for x < 0, let me set x = -t where t > 0, then x + 1/x = -t - 1/t = - (t + 1/t) ≤ -2. Therefore, the range of y is (-∞, -2] ∪ [2, ∞).

Therefore, for the original equation to have a real root, the quadratic equation in y must have at least one root in (-∞, -2] ∪ [2, ∞).

Therefore, the problem reduces to finding the set of (a, b) in [0,1]x[0,1] such that the quadratic equation y² + a y - (b + 2) = 0 has at least one root in (-∞, -2] ∪ [2, ∞).

So, now, we need to find the conditions on a and b such that this quadratic equation has at least one root in (-∞, -2] ∪ [2, ∞).

Let me denote the quadratic equation as:

y² + a y - (b + 2) = 0.

Let me denote this as f(y) = y² + a y - (b + 2).

We need f(y) = 0 to have at least one root in (-∞, -2] ∪ [2, ∞).

To find this, we can analyze the quadratic function f(y). Since it's a quadratic opening upwards (coefficient of y² is 1 > 0), the graph is a parabola opening upwards. Therefore, the equation f(y) = 0 will have roots in (-∞, -2] ∪ [2, ∞) if either:

1. The smaller root is ≤ -2, or

2. The larger root is ≥ 2.

But since the parabola opens upwards, if the smaller root is ≤ -2, then the entire left side of the parabola will be negative up to that root, but since it's opening upwards, the function tends to +infty as y approaches ±infty. Wait, maybe I need to think more carefully.

Alternatively, for the quadratic equation to have at least one root in (-∞, -2] ∪ [2, ∞), we can use the following approach:

Either:

- The quadratic has a root ≥ 2, or

- The quadratic has a root ≤ -2.

But since the quadratic is opening upwards, if there is a root ≥ 2, then f(2) ≤ 0, because the parabola crosses the y-axis at f(2) and since it opens upwards, if there is a root ≥ 2, then f(2) ≤ 0. Similarly, if there is a root ≤ -2, then f(-2) ≤ 0.

Wait, let me recall the concept here. For a quadratic equation f(y) = 0 with leading coefficient positive, if there is a root ≥ k, then f(k) ≤ 0. Similarly, if there is a root ≤ m, then f(m) ≤ 0. Because if the parabola crosses the x-axis at some point ≥ k, then at y = k, the function is below or equal to zero. Similarly for the other side.

Therefore, the condition that the quadratic has at least one root in (-∞, -2] ∪ [2, ∞) is equivalent to f(-2) ≤ 0 or f(2) ≤ 0. Wait, but is that sufficient?

Wait, suppose f(-2) ≤ 0. Then since the parabola opens upwards, there must be a root ≤ -2? Wait, not necessarily. Because if f(-2) ≤ 0, then since the parabola opens upwards, there is a root to the left of -2 or between -2 and the other root. Wait, maybe I need to think more carefully.

Let me recall that for a quadratic equation f(y) = 0 with leading coefficient positive, the following holds:

- If f(k) ≤ 0, then there is at least one root in (-∞, k] if the parabola crosses the x-axis to the left of k. Wait, no. Actually, if f(k) ≤ 0, then since the parabola opens upwards, there must be at least one root ≤ k and another root ≥ k? No, that's not necessarily true. For example, if the entire parabola is below zero at y = k, but since it opens upwards, it must cross the x-axis on both sides of k if there are two real roots. Wait, but if f(k) ≤ 0, then there must be at least one root ≤ k and another root ≥ k? Wait, no. Suppose the quadratic has two real roots. If f(k) ≤ 0, then k lies between the two roots. Because if the parabola opens upwards, then between the two roots, the function is negative. So if f(k) ≤ 0, then k is between the two roots. Therefore, if f(-2) ≤ 0, then -2 is between the two roots, meaning that one root is ≤ -2 and the other is ≥ -2. Similarly, if f(2) ≤ 0, then 2 is between the two roots, meaning one root is ≤ 2 and the other is ≥ 2.

But we need at least one root in (-∞, -2] ∪ [2, ∞). So, if f(-2) ≤ 0, then there is a root ≤ -2 (since one root is ≤ -2 and the other is ≥ -2). Similarly, if f(2) ≤ 0, then there is a root ≥ 2. Therefore, the condition that the quadratic has at least one root in (-∞, -2] ∪ [2, ∞) is equivalent to f(-2) ≤ 0 or f(2) ≤ 0. Wait, but is that correct?

Wait, suppose f(-2) ≤ 0. Then, since the parabola opens upwards, there are two real roots, one ≤ -2 and one ≥ -2. Therefore, there is a root ≤ -2. Similarly, if f(2) ≤ 0, then there are two real roots, one ≤ 2 and one ≥ 2, so there is a root ≥ 2. Therefore, the condition that the quadratic has at least one root in (-∞, -2] ∪ [2, ∞) is equivalent to f(-2) ≤ 0 or f(2) ≤ 0. However, we also need to check if there are cases where both f(-2) > 0 and f(2) > 0, but the quadratic still has a root in (-∞, -2] ∪ [2, ∞). Is that possible?

Wait, suppose that the quadratic has two real roots, both greater than 2. Then f(2) would be positive, because the parabola is opening upwards, so at y=2, which is to the left of both roots, the function is positive. Similarly, if both roots are less than -2, then f(-2) would be positive. Therefore, if the quadratic has both roots in (-∞, -2] ∪ [2, ∞), then f(-2) and f(2) would be positive? Wait, no. If both roots are greater than 2, then f(2) is positive. If both roots are less than -2, then f(-2) is positive. However, if one root is less than -2 and the other is greater than 2, then f(-2) and f(2) would be negative? Wait, let me think.

Suppose the quadratic has one root less than -2 and one root greater than 2. Then, since the parabola opens upwards, at y = -2, which is between the two roots, the function f(-2) would be negative. Similarly, at y = 2, which is between the two roots, f(2) would be negative. Therefore, in this case, both f(-2) and f(2) are negative. Therefore, if the quadratic has one root in (-∞, -2] and one root in [2, ∞), then both f(-2) ≤ 0 and f(2) ≤ 0. However, if the quadratic has both roots in (-∞, -2], then f(-2) would be positive (since the parabola is above zero at y=-2, as both roots are to the left of -2). Similarly, if both roots are in [2, ∞), then f(2) is positive. Therefore, the condition that the quadratic has at least one root in (-∞, -2] ∪ [2, ∞) is equivalent to f(-2) ≤ 0 or f(2) ≤ 0. Because if f(-2) ≤ 0, then there is a root ≤ -2, and if f(2) ≤ 0, there is a root ≥ 2. However, if both f(-2) ≤ 0 and f(2) ≤ 0, then there are roots in both intervals. Therefore, the set of (a, b) such that the original equation has a real root is the set of (a, b) where f(-2) ≤ 0 or f(2) ≤ 0.

Therefore, we can write:

Either f(-2) ≤ 0 or f(2) ≤ 0.

So let's compute f(-2) and f(2):

First, f(y) = y² + a y - (b + 2).

Therefore:

f(-2) = (-2)² + a*(-2) - (b + 2) = 4 - 2a - b - 2 = 2 - 2a - b.

Similarly, f(2) = (2)² + a*(2) - (b + 2) = 4 + 2a - b - 2 = 2 + 2a - b.

Therefore, the conditions are:

Either 2 - 2a - b ≤ 0 or 2 + 2a - b ≤ 0.

So, simplifying:

First condition: 2 - 2a - b ≤ 0 ⇒ 2a + b ≥ 2.

Second condition: 2 + 2a - b ≤ 0 ⇒ 2a - b ≤ -2 ⇒ b ≥ 2a + 2.

But since a and b are between 0 and 1, let's analyze these inequalities.

First, for the first condition: 2a + b ≥ 2.

Given that a and b are in [0,1], let's see what region this defines.

Similarly, for the second condition: b ≥ 2a + 2. However, since b ≤ 1, and 2a + 2 ≥ 2*0 + 2 = 2, but b ≤ 1 < 2, so this inequality b ≥ 2a + 2 can never be satisfied. Therefore, the second condition is impossible. Therefore, the only condition we need is 2a + b ≥ 2.

Wait, that seems important. Let me check again.

Second condition: 2 + 2a - b ≤ 0 ⇒ 2a - b ≤ -2 ⇒ b ≥ 2a + 2. Since a is between 0 and 1, 2a is between 0 and 2, so 2a + 2 is between 2 and 4. However, since b is between 0 and 1, b cannot be ≥ 2a + 2, because 2a + 2 ≥ 2 > 1. Therefore, the second condition is impossible. Therefore, the only condition is 2a + b ≥ 2.

Therefore, the set S is the set of (a, b) in [0,1]x[0,1] such that 2a + b ≥ 2.

Wait, but hold on. Let me confirm this conclusion. Because if the second condition is impossible, then the only way for the quadratic to have a root in (-∞, -2] ∪ [2, ∞) is if 2a + b ≥ 2. Therefore, the set S is the region in the unit square where 2a + b ≥ 2. Therefore, the area of S is the area of the region defined by 2a + b ≥ 2 with 0 ≤ a, b ≤ 1.

But let me check if this is correct. Let me think again. Because earlier, I thought that if f(-2) ≤ 0 or f(2) ≤ 0, but since f(2) ≤ 0 is impossible, only f(-2) ≤ 0 is possible. Therefore, the condition reduces to 2 - 2a - b ≤ 0 ⇒ 2a + b ≥ 2. Therefore, the set S is the set of (a, b) with 2a + b ≥ 2, 0 ≤ a, b ≤ 1.

But let me check with an example. Let me take a = 1, b = 0. Then 2a + b = 2*1 + 0 = 2, which satisfies the condition. Let me check if the original equation has a real root. The equation becomes x⁴ + x³ - 0x² + x + 1 = x⁴ + x³ + x + 1. Let me try to see if this has real roots. Let me plug in x = -1: (-1)^4 + (-1)^3 + (-1) + 1 = 1 - 1 - 1 + 1 = 0. So x = -1 is a root. Therefore, yes, there is a real root. So that point (1, 0) is in S.

Another example: take a = 0.5, b = 1. Then 2a + b = 1 + 1 = 2, which satisfies the condition. Let me check the equation: x⁴ + 0.5x³ - 1x² + 0.5x + 1. Let me try x = -1: 1 - 0.5 - 1 - 0.5 + 1 = 0. So again, x = -1 is a root. So that works.

Another example: take a = 0. Let me see. If a = 0, then 2a + b = b. So to satisfy 2a + b ≥ 2, we need b ≥ 2. But since b ≤ 1, there are no solutions here. Therefore, when a = 0, there are no points in S. Which makes sense, because if a = 0, the equation becomes x⁴ - b x² + 1 = 0. Let me check if this has real roots. Let me set y = x², then equation becomes y² - b y + 1 = 0. The discriminant is b² - 4. Since b ≤ 1, discriminant is ≤ 1 - 4 = -3 < 0. Therefore, no real roots. Therefore, when a = 0, there are no real roots, which matches our previous conclusion.

Another example: take a = 1, b = 1. Then 2a + b = 2 + 1 = 3 ≥ 2. Let me check the equation: x⁴ + x³ - x² + x + 1. Let me try x = -1: 1 - 1 - 1 - 1 + 1 = -1 ≠ 0. Hmm, so x = -1 is not a root here. Wait, but according to our previous logic, since 2a + b ≥ 2, there should be a real root. Maybe I need to check another value. Let me try x = -2: (-2)^4 + (-2)^3 - (-2)^2 + (-2) + 1 = 16 - 8 - 4 - 2 + 1 = 3 ≠ 0. Hmm. Maybe there's another real root?

Alternatively, maybe I need to check the transformed equation. Let me use the substitution y = x + 1/x. For the equation x⁴ + x³ - x² + x + 1 = 0. Dividing by x²: x² + x - 1 + 1/x + 1/x² = 0. Which is y² - 2 + a y - b = 0? Wait, earlier we had y² + a y - (b + 2) = 0. Wait, in this case, a = 1, b = 1. Therefore, the equation is y² + y - (1 + 2) = y² + y - 3 = 0. Solving this quadratic: y = [-1 ± √(1 + 12)] / 2 = [-1 ± √13]/2. So the roots are approximately [-1 + 3.605]/2 ≈ 1.3025 and [-1 - 3.605]/2 ≈ -2.3025. Now, since y must be in (-∞, -2] ∪ [2, ∞). The first root is ≈1.3025 which is between -2 and 2, so not in the required range. The second root is ≈-2.3025 which is less than -2. Therefore, there is a real root corresponding to y ≈ -2.3025. Therefore, there exists real x such that x + 1/x ≈ -2.3025. Let me check if such x exists. For example, solving x + 1/x = -2.3025. Let me set z = x, then z + 1/z = -k where k > 0. Multiply both sides by z: z² + 1 = -k z ⇒ z² + k z + 1 = 0. The discriminant is k² - 4. For real solutions, we need k² - 4 ≥ 0 ⇒ k ≥ 2. In our case, k ≈ 2.3025, which is greater than 2, so there are real solutions. Therefore, there are real roots. Therefore, even though x = -1 is not a root, there are other real roots. Therefore, the point (1, 1) is indeed in S. Therefore, our condition seems correct.

Therefore, the set S is defined by 2a + b ≥ 2 with 0 ≤ a, b ≤ 1. Therefore, the area of S is the area of the region in the unit square where 2a + b ≥ 2.

Now, let me visualize this region. The line 2a + b = 2 in the a-b plane (with a and b between 0 and 1). Let me find the intercepts.

When a = 0: 0 + b = 2 ⇒ b = 2, but since b ≤ 1, this is outside the square.

When b = 0: 2a + 0 = 2 ⇒ a = 1. So the line passes through (1, 0).

When a = 1: 2*1 + b = 2 ⇒ b = 0. So the line passes through (1, 0) and (0, 2), but since b can't be 2, the line within the square goes from (1, 0) to where? Let me find another point. Let me solve for when b = 1: 2a + 1 = 2 ⇒ 2a = 1 ⇒ a = 0.5. Therefore, the line passes through (0.5, 1). Therefore, within the unit square, the line 2a + b = 2 connects (1, 0) to (0.5, 1). Therefore, the region 2a + b ≥ 2 is the area above this line within the square.

Therefore, the area of S is the area above the line from (1, 0) to (0.5, 1) in the unit square.

To compute this area, we can compute the area of the region in the unit square where 2a + b ≥ 2. This is a triangle or a trapezoid?

Let me sketch mentally. The line goes from (1, 0) to (0.5, 1). The region 2a + b ≥ 2 is above this line. So, in the unit square, the area above the line is a triangle with vertices at (1, 0), (0.5, 1), and (0, 1)? Wait, no. Wait, when a = 0, b must be ≥ 2, which is outside the square. So the region within the square is bounded by the line from (1, 0) to (0.5, 1), and then the top edge of the square from (0.5, 1) to (0, 1), and the left edge from (0, 1) to (0, 0)? Wait, no. Wait, actually, the region 2a + b ≥ 2 in the square is a polygon with vertices at (1, 0), (0.5, 1), and (0, 1)? Wait, let me check.

At b = 1, the line 2a + 1 = 2 ⇒ a = 0.5. So the line intersects the top edge of the square (b=1) at (0.5, 1). Then, if we go to the left, when a decreases, b would have to increase to satisfy 2a + b ≥ 2, but since b can't exceed 1, the region is bounded by the line from (1, 0) to (0.5, 1), and then the line from (0.5, 1) to (0, 1) along b=1? Wait, but for a < 0.5, even if b=1, 2a + 1 ≥ 2 ⇒ 2a ≥ 1 ⇒ a ≥ 0.5. Therefore, for a < 0.5, even with b=1, 2a + b < 2. Therefore, the region 2a + b ≥ 2 is only above the line from (1, 0) to (0.5, 1), and for a between 0.5 and 1, b between 2 - 2a and 1. Therefore, the area is a triangle with vertices at (1, 0), (0.5, 1), and (1, 1). Wait, let me check.

Wait, when a = 1, b can be from 0 to 1, but 2a + b ≥ 2 ⇒ 2*1 + b ≥ 2 ⇒ b ≥ 0. So for a = 1, all b between 0 and 1 satisfy the condition. Wait, but earlier when we checked a = 1, b = 0, it worked. But when a = 1, b can be from 0 to 1. However, according to the line 2a + b = 2, when a = 1, b = 0. So the region for a = 1 is from b = 0 to 1. Similarly, for a between 0.5 and 1, the lower bound for b is 2 - 2a, and upper bound is 1. For a between 0 and 0.5, there is no b in [0,1] that satisfies 2a + b ≥ 2, since 2a ≤ 1, so 2a + b ≤ 1 + 1 = 2, but equality only when a = 0.5 and b = 1. Therefore, the region is a trapezoid? Wait, no. Wait, for a from 0.5 to 1, the region is between b = 2 - 2a and b = 1. So the area can be calculated as the integral from a = 0.5 to 1 of [1 - (2 - 2a)] da. Which is integral from 0.5 to 1 of (2a - 1) da.

Alternatively, since it's a triangle. Let me think. The region is a triangle with vertices at (1, 0), (1, 1), and (0.5, 1). Because when a = 1, b ranges from 0 to 1, and when a = 0.5, b ranges from 1 to 1 (i.e., just the point (0.5, 1)). Therefore, the area is the area of the triangle with vertices at (1, 0), (1, 1), and (0.5, 1). The base of this triangle is from (1, 0) to (1, 1), which is length 1. The height is from (1, 1) to (0.5, 1), which is 0.5. But since it's a right triangle? Wait, no. Wait, the three points are (1, 0), (1, 1), and (0.5, 1). So plotting these points, it's a right triangle with legs of length 1 (from (1,0) to (1,1)) and 0.5 (from (1,1) to (0.5,1)). Therefore, the area is (base * height)/2 = (1 * 0.5)/2 = 0.25. Wait, but let me check with integration.

Compute the area as integral from a = 0.5 to 1 of [upper bound - lower bound] da. Upper bound is 1, lower bound is 2 - 2a. Therefore, the integral is ∫_{0.5}^1 [1 - (2 - 2a)] da = ∫_{0.5}^1 (2a - 1) da.

Compute the integral:

∫ (2a - 1) da = a² - a evaluated from 0.5 to 1.

At a = 1: 1² - 1 = 0.

At a = 0.5: (0.5)² - 0.5 = 0.25 - 0.5 = -0.25.

Therefore, the integral is 0 - (-0.25) = 0.25. Therefore, the area is 0.25. Therefore, the area of S is 1/4.

Wait, but let me think again. If the region is a triangle with vertices at (1, 0), (1, 1), and (0.5, 1), then the area is indeed 1/2 * base * height. The base can be considered as the vertical segment from (1, 0) to (1, 1), which is length 1. The height is the horizontal distance from this base to the point (0.5, 1), which is 0.5. However, since it's a right triangle, the area is (base * height)/2 = (1 * 0.5)/2 = 0.25. So that's 1/4. Therefore, the area is 1/4.

But wait, let me check with another approach. The line 2a + b = 2 cuts the square. The area above the line is a triangle with area 1/4. Therefore, the answer is 1/4? But let me check with another example. Suppose we have the unit square [0,1]x[0,1]. The line 2a + b = 2 goes from (1, 0) to (0.5, 1). The area above this line is the triangle we discussed. Alternatively, the area below the line is the rest of the square. The area below the line would be the area of the trapezoid from (0,0) to (1,0) to (0.5,1) to (0,1). Wait, but maybe not. Let me compute the area below the line 2a + b = 2 in the square. For a from 0 to 0.5, b can go up to 1, but the line 2a + b = 2 would have b = 2 - 2a. However, for a from 0 to 0.5, 2 - 2a ranges from 2 to 1. Therefore, the area below the line is the area under the line from a=0 to a=0.5, plus the area from a=0.5 to a=1, where the line is above b=1. Wait, this is getting complicated. Alternatively, since the total area of the square is 1, and the area above the line is 1/4, then the area below the line would be 3/4. But let me check with integrating.

Alternatively, the area above the line 2a + b ≥ 2 is the integral from a=0.5 to 1 of (1 - (2 - 2a)) da = 0.25 as before. Therefore, the area is 1/4. Therefore, the answer is 1/4? But let me check with another method.

Alternatively, think of the region 2a + b ≥ 2 in the unit square. This is equivalent to b ≥ 2 - 2a. So for each a between 0.5 and 1, b ranges from 2 - 2a to 1. For a between 0 and 0.5, there is no b in [0,1] that satisfies b ≥ 2 - 2a, since 2 - 2a ≥ 1 when a ≤ 0.5. Therefore, the area is the integral from a=0.5 to 1 of [1 - (2 - 2a)] da = integral from 0.5 to 1 of (2a - 1) da = [a² - a] from 0.5 to 1 = (1 - 1) - (0.25 - 0.5) = 0 - (-0.25) = 0.25. Therefore, 0.25 is 1/4. Therefore, the area is 1/4.

But wait, the problem says "the set of points (a,b) with 0 ≤ a, b ≤ 1 such that the equation... has at least one real root." So according to our analysis, this area is 1/4. However, let me check again if there are other cases where the equation has real roots. Earlier, we considered only the case where the quadratic in y has a root in (-∞, -2] ∪ [2, ∞). But is there a possibility that even if f(-2) > 0 and f(2) > 0, the quadratic could still have a root in (-∞, -2] ∪ [2, ∞)? For example, if the quadratic has two real roots, both outside of (-2, 2). Wait, but if the quadratic has two real roots, then since the parabola opens upwards, if both roots are greater than 2, then f(2) > 0. Similarly, if both roots are less than -2, then f(-2) > 0. Therefore, if the quadratic has two real roots outside of (-2, 2), then both f(-2) and f(2) are positive. Therefore, in that case, the equation would have real roots, but our condition f(-2) ≤ 0 or f(2) ≤ 0 would not be satisfied. Therefore, our previous conclusion that the condition is equivalent to f(-2) ≤ 0 or f(2) ≤ 0 is incomplete? Wait, but earlier we thought that if the quadratic has a root in (-∞, -2] ∪ [2, ∞), then either f(-2) ≤ 0 or f(2) ≤ 0. But if the quadratic has two roots outside of (-2, 2), then both f(-2) and f(2) are positive. Therefore, our previous conclusion was wrong? Wait, this seems contradictory. Let me think again.

Let me take an example where the quadratic has two roots outside of (-2, 2). For example, suppose the quadratic equation y² + a y - (b + 2) = 0 has two roots both greater than 2. Then, since the parabola opens upwards, f(2) = 2 + 2a - b. If both roots are greater than 2, then f(2) must be positive. Similarly, if both roots are less than -2, then f(-2) is positive. Therefore, in such cases, the quadratic does have real roots outside of (-2, 2), but our condition f(-2) ≤ 0 or f(2) ≤ 0 is not satisfied. Therefore, our previous conclusion was wrong. Therefore, there is a mistake in the reasoning.

Wait, so where is the mistake? Earlier, I thought that if the quadratic has a root in (-∞, -2] ∪ [2, ∞), then f(-2) ≤ 0 or f(2) ≤ 0. But actually, if the quadratic has two roots outside of (-2, 2), then both f(-2) and f(2) are positive. Therefore, the condition that the quadratic has at least one root in (-∞, -2] ∪ [2, ∞) is not equivalent to f(-2) ≤ 0 or f(2) ≤ 0. Therefore, my previous conclusion was wrong. Therefore, I need to re-examine the problem.

So, going back. The quadratic equation y² + a y - (b + 2) = 0 has real roots if discriminant is non-negative, which it always is. The roots are y = [-a ± √(a² + 4(b + 2))]/2. We need at least one of these roots to be in (-∞, -2] ∪ [2, ∞). Therefore, we need to find conditions on a and b such that at least one of the roots is in that interval.

To find this, we can use the following approach: For a quadratic equation f(y) = 0, the condition that at least one root is ≥ 2 is equivalent to f(2) ≤ 0 or the quadratic has a root ≥ 2. Similarly for ≤ -2. However, since the quadratic opens upwards, the condition that there is a root ≥ 2 is equivalent to f(2) ≤ 0 or the vertex is to the right of 2 and the minimum value is ≤ 0. Wait, this is getting complicated. Maybe another approach is better.

Alternatively, for the quadratic equation y² + a y - (b + 2) = 0, let me denote the roots as y₁ and y₂. We need at least one of y₁ or y₂ to be ≤ -2 or ≥ 2.

This can be approached by considering the following cases:

Case 1: At least one root ≥ 2.

Case 2: At least one root ≤ -2.

But these cases can overlap. However, since the quadratic is symmetric in some way, maybe we can find conditions for each case.

For Case 1: At least one root ≥ 2.

This occurs if either:

- The quadratic has a root exactly at 2, or

- The quadratic has a root greater than 2.

For a quadratic with leading coefficient positive, this happens if f(2) ≤ 0 or the vertex is to the right of 2 and the minimum value is ≤ 0. Wait, but since the parabola opens upwards, if f(2) ≤ 0, then there is a root ≥ 2. If f(2) > 0, then both roots are either less than 2 or one less and one greater. Wait, no. If f(2) > 0, then since the parabola opens upwards, if there are two real roots, they are both on the same side of 2? No, if f(2) > 0, then depending on the position of the vertex, the roots could be both less than 2 or both greater than 2? Wait, no. If the parabola opens upwards and f(2) > 0, then if there are two real roots, they are either both less than 2 or both greater than 2? No, that's not necessarily true. For example, take f(y) = y² - 4. Then f(2) = 0, which is the case of a root at 2. If f(y) = y² - 3, then f(2) = 4 - 3 = 1 > 0, and the roots are ±√3 ≈ ±1.732, which are both less than 2. If f(y) = y² - 5, then f(2) = 4 - 5 = -1 < 0, so there are roots on both sides of 2. Wait, no. Wait, if f(2) < 0, then since the parabola opens upwards, there are two real roots, one less than 2 and one greater than 2. If f(2) = 0, then one root is at 2. If f(2) > 0, then either both roots are less than 2 or both are greater than 2? Wait, no. If f(2) > 0, then depending on the vertex, it could be that both roots are less than 2 or both are greater than 2. For example, take f(y) = y² - 3, which has roots at ±√3 ≈ ±1.732. So both roots are less than 2. If we take f(y) = y² - 5, which has roots at ±√5 ≈ ±2.236. So one root is less than 2 and one is greater than 2. Wait, no, √5 ≈ 2.236 > 2, and -√5 ≈ -2.236 < -2. So in this case, one root is less than -2, and one is greater than 2. Wait, but f(2) = 4 - 5 = -1 < 0. So if f(2) < 0, then there is one root less than 2 and one greater than 2. If f(2) > 0, then both roots are either less than 2 or both greater than 2? Wait, no. For example, take f(y) = y² - 4y + 3. This factors as (y - 1)(y - 3). So roots at 1 and 3. f(2) = 4 - 8 + 3 = -1 < 0. So here, f(2) < 0, and there are roots on both sides of 2. If I take f(y) = y² - 4y + 5. Then discriminant is 16 - 20 = -4 < 0, no real roots. If I take f(y) = y² - 4y + 4 = (y - 2)^2. Then f(2) = 0, and there's a double root at 2. If I take f(y) = y² - 4y + 6. Discriminant is 16 - 24 = -8 < 0. No real roots. If I take f(y) = y² - 4y + 2. Discriminant is 16 - 8 = 8 > 0. Roots at [4 ± √8]/2 = 2 ± √2 ≈ 2 ± 1.414. So one root ≈ 3.414 > 2, and one root ≈ 0.586 < 2. Therefore, f(2) = 4 - 8 + 2 = -2 < 0. So again, when f(2) < 0, there are roots on both sides of 2. If f(2) > 0, then there are no real roots or both roots on one side. Wait, but if f(2) > 0, and there are real roots, then they must be both on one side of 2. For example, take f(y) = y² - 3, which has roots at ±√3 ≈ ±1.732. Both less than 2. If I take f(y) = y² - 5, which has roots at ±√5 ≈ ±2.236. So one root is less than -2, and one is greater than 2. Wait, but f(2) = 4 - 5 = -1 < 0. So in this case, f(2) < 0. Therefore, if f(2) > 0, then there are no real roots or both roots are on the same side of 2. Wait, but how can there be real roots with f(2) > 0 and both roots on the same side of 2? For example, take f(y) = y² - 4y + 5. Wait, discriminant is negative. How about f(y) = y² - 4y + 6. Discriminant negative. How about f(y) = y² - 4y + 4. Then f(2) = 0, double root at 2. If I take f(y) = y² - 4y + 3. Then f(2) = -1 < 0. So maybe if f(2) > 0, then there are no real roots or both roots are on the same side of 2. Wait, but how can there be real roots with f(2) > 0 and both roots on the same side of 2? For example, take f(y) = y² - 3. Then f(2) = 4 - 3 = 1 > 0, and roots at ±√3 ≈ ±1.732. Both roots are less than 2. So yes, in this case, both roots are less than 2. Similarly, if I take f(y) = y² - 5, then f(2) = 4 - 5 = -1 < 0, and roots at ±√5 ≈ ±2.236. So one root is less than -2, one is greater than 2. Therefore, if f(2) > 0, then both roots are less than 2 or both are greater than 2? Wait, no. If f(2) > 0, and there are real roots, then they are both less than 2 or both greater than 2? How can they be both greater than 2? For example, take f(y) = y² - 4y + 5. Wait, discriminant is negative. How about f(y) = y² - 4y + 6. Still discriminant negative. How about f(y) = y² - 4y + 7. Still discriminant negative. Wait, maybe it's not possible to have both roots greater than 2 with f(2) > 0? Let me think. Suppose we have a quadratic with both roots greater than 2. Then, since the parabola opens upwards, f(2) would be positive, because at y = 2, which is to the left of both roots, the function is positive. Therefore, yes, if both roots are greater than 2, then f(2) > 0. Similarly, if both roots are less than 2, then f(2) > 0. Therefore, the condition that there is at least one root ≥ 2 is equivalent to f(2) ≤ 0 or the quadratic has both roots ≥ 2. Wait, but how do we check for both roots ≥ 2?

For a quadratic to have both roots ≥ 2, the following conditions must be satisfied:

1. The quadratic has real roots (discriminant ≥ 0).

2. The axis of symmetry is ≥ 2.

3. f(2) ≥ 0.

Wait, the axis of symmetry is at y = -a/2. For both roots to be ≥ 2, the axis of symmetry must be ≥ 2, and f(2) ≥ 0. Wait, no. Let me recall the conditions for both roots ≥ k:

For a quadratic f(y) = y² + py + q, to have both roots ≥ k, the following must hold:

1. Discriminant ≥ 0.

2. The axis of symmetry ≥ k.

3. f(k) ≥ 0.

Similarly, for both roots ≤ k, the axis of symmetry ≤ k and f(k) ≥ 0.

Therefore, applying this to our case, for both roots ≥ 2:

1. Discriminant ≥ 0 (which it always is).

2. Axis of symmetry ≥ 2. The axis of symmetry is at y = -a/2. So -a/2 ≥ 2 ⇒ -a ≥ 4 ⇒ a ≤ -4. But since a is between 0 and 1, this is impossible. Therefore, there are no such quadratics with both roots ≥ 2. Therefore, the only way to have at least one root ≥ 2 is if f(2) ≤ 0. Similarly, for at least one root ≤ -2, we need f(-2) ≤ 0. Therefore, our original conclusion was correct! Because for the quadratic to have at least one root ≥ 2, since the axis of symmetry cannot be ≥ 2 (as a ≥ 0), the only possibility is f(2) ≤ 0. Similarly, for at least one root ≤ -2, since the axis of symmetry is at y = -a/2, which for a ≥ 0 is ≤ 0, so to have a root ≤ -2, we need f(-2) ≤ 0. Therefore, the condition that the quadratic has at least one root in (-∞, -2] ∪ [2, ∞) is equivalent to f(-2) ≤ 0 or f(2) ≤ 0. And since f(2) ≤ 0 is equivalent to 2 + 2a - b ≤ 0 ⇒ b ≥ 2a + 2, which is impossible, and f(-2) ≤ 0 is equivalent to 2 - 2a - b ≤ 0 ⇒ 2a + b ≥ 2. Therefore, the only condition is 2a + b ≥ 2. Therefore, the area is indeed 1/4.

But earlier, I thought there was a contradiction, but now it seems that the previous conclusion was correct. The confusion was due to thinking that there could be cases where both roots are outside of (-2, 2), but in reality, for our specific quadratic, since the axis of symmetry is at y = -a/2, which for a ≥ 0 is ≤ 0, so the quadratic cannot have both roots ≥ 2. Therefore, the only way to have a root ≥ 2 is f(2) ≤ 0, which is impossible, and the only way to have a root ≤ -2 is f(-2) ≤ 0. Therefore, the condition is indeed 2a + b ≥ 2, and the area is 1/4.

Therefore, the answer should be 1/4, so the area of S is 1/4, which is 0.25. Therefore, the final answer is \boxed{\dfrac{1}{4}}.

**Final Answer**
\boxed{\dfrac{1}{4}}

To determine the area of the set $ S $, we are given the condition that the quartic polynomial

$$
x^4 + ax^3 - bx^2 + ax + 1 = 0
$$

has at least one real root, for $ (a, b) \in [0,1] \times [0,1] $. Our goal is to find the area of the region in the unit square where this condition is satisfied.

---

### Step 1: Recognize the Polynomial Structure

The polynomial is symmetric in a special way. It is **palindromic**, meaning its coefficients read the same forwards and backwards. This allows us to use a substitution to simplify the equation.

Divide the equation by $ x^2 $ (valid for $ x \ne 0 $):

$$
x^2 + a x - b + \frac{a}{x} + \frac{1}{x^2} = 0
$$

Let $ y = x + \frac{1}{x} $. Then, we know:

$$
x^2 + \frac{1}{x^2} = y^2 - 2
$$

Substituting into the equation:

$$
y^2 - 2 + a y - b = 0 \quad \Rightarrow \quad y^2 + a y - (b + 2) = 0
$$

This is a quadratic in $ y $, and we are interested in whether this equation has a real solution $ y $ such that $ y = x + \frac{1}{x} $ for some real $ x \ne 0 $.

---

### Step 2: Determine the Range of $ y = x + \frac{1}{x} $

For real $ x \ne 0 $, the expression $ x + \frac{1}{x} $ has the range:

$$
(-\infty, -2] \cup [2, \infty)
$$

This is because:

- For $ x > 0 $, by AM-GM, $ x + \frac{1}{x} \ge 2 $
- For $ x < 0 $, $ x + \frac{1}{x} \le -2 $

So, the original equation has a real root if and only if the quadratic in $ y $ has at least one root in $ (-\infty, -2] \cup [2, \infty) $.

---

### Step 3: Analyze the Quadratic in $ y $

Let the quadratic be:

$$
f(y) = y^2 + a y - (b + 2)
$$

We want $ f(y) = 0 $ to have at least one root in $ (-\infty, -2] \cup [2, \infty) $.

Since the parabola opens upwards (coefficient of $ y^2 $ is positive), the condition for having a root in $ (-\infty, -2] \cup [2, \infty) $ is equivalent to:

- $ f(-2) \le 0 $ or $ f(2) \le 0 $

Let’s compute these:

- $ f(-2) = (-2)^2 + a(-2) - (b + 2) = 4 - 2a - b - 2 = 2 - 2a - b $
- $ f(2) = 2^2 + a(2) - (b + 2) = 4 + 2a - b - 2 = 2 + 2a - b $

So the conditions become:

- $ 2 - 2a - b \le 0 \Rightarrow 2a + b \ge 2 $
- $ 2 + 2a - b \le 0 \Rightarrow b \ge 2a + 2 $

Now, since $ a, b \in [0,1] $, the second inequality $ b \ge 2a + 2 $ is **never satisfied**, because the right-hand side is at least 2, while $ b \le 1 $. So the only valid condition is:

$$
2a + b \ge 2
$$

---

### Step 4: Determine the Region in the Unit Square

We now find the area of the region in the unit square $ [0,1] \times [0,1] $ where $ 2a + b \ge 2 $.

This is a linear inequality. The line $ 2a + b = 2 $ intersects the square at:

- When $ a = 0 $: $ b = 2 $ → outside the square
- When $ b = 0 $: $ a = 1 $
- When $ b = 1 $: $ 2a + 1 = 2 \Rightarrow a = 0.5 $

So the line passes through the points $ (1, 0) $ and $ (0.5, 1) $, and the region $ 2a + b \ge 2 $ lies **above** this line within the square.

This region is a **right triangle** with vertices at:

- $ (1, 0) $
- $ (1, 1) $
- $ (0.5, 1) $

The base of the triangle is from $ (1, 0) $ to $ (1, 1) $, which has length 1.

The height is from $ (1, 1) $ to $ (0.5, 1) $, which has length $ 0.5 $.

So the area is:

$$
\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 1 \times 0.5 = \frac{1}{4}
$$

---

### Final Answer

$$
\boxed{\dfrac{1}{4}}
$$

Solution:
Note that $x = 0$ cannot be a solution of the equation.  Dividing both sides by $x^2,$ we get
\[x^2 + ax - b + \frac{a}{x} + \frac{1}{x^2} = 0.\]Let $y = x + \frac{1}{x}.$  Then $x^2 - yx + 1 = 0.$  The discriminant of this quadratic is
\[y^2 - 4,\]so there is a real root in $x$ as long as $|y| \ge 2.$

Also, $y^2 = x^2 + 2 + \frac{1}{x^2},$ so
\[y^2 + ay - (b + 2) = 0.\]By the quadratic formula, the roots are
\[y = \frac{-a \pm \sqrt{a^2 + 4(b + 2)}}{2}.\]First, we notice that the discriminant $a^2 + 4(b + 2)$ is always positive.  Furthermore, there is a value $y$ such that $|y| \ge 2$ as long as
\[\frac{a + \sqrt{a^2 + 4(b + 2)}}{2} \ge 2.\]Then $a + \sqrt{a^2 + 4(b + 2)} \ge 4,$ or $\sqrt{a^2 + 4(b + 2)} \ge 4 - a.$  Both sides are nonnegative, so we can square both sides, to get
\[a^2 + 4(b + 2) \ge a^2 - 8a + 16.\]This simplifies to $2a + b \ge 2.$

[asy]
unitsize(3 cm);

fill((1/2,1)--(1,0)--(1,1)--cycle,gray(0.7));
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
draw((1/2,1)--(1,0));

label("$0$", (0,0), S);
label("$1$", (1,0), S);
label("$a$", (1,0), E);
label("$0$", (0,0), W);
label("$1$", (0,1), W);
label("$b$", (0,1), N);
[/asy]

Thus, $S$ is the triangle whose vertices are $(1,0),$ $(1,1),$ and $\left( \frac{1}{2}, 1 \right),$ which has area $\boxed{\frac{1}{4}}.$