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 one root in (-∞, -2] and another root in [2, ∞). Then, since the parabola opens upwards, f(-2) would be negative (since between the two roots, the function is negative). Similarly, f(2) would be negative. Therefore, if there is a root in (-∞, -2] and another in [2, ∞), then both f(-2) ≤ 0 and 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. Wait, but if f(-2) ≤ 0 or f(2) ≤ 0, then there is at least one root in the required intervals. 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 union of the sets where f(-2) ≤ 0 or f(2) ≤ 0.

Therefore, we can write:

Original equation has real root ⇨ f(-2) ≤ 0 or f(2) ≤ 0.

Therefore, let's compute f(-2) and f(2):

First, f(y) = y² + a y - (b + 2).

Compute f(-2):

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

Therefore, we need to find the area of the region in the unit square [0,1]x[0,1] where 2 - 2a - b ≤ 0 or 2 + 2a - b ≤ 0.

Let me write these inequalities:

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

Second inequality: 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 second inequality: b ≥ 2a + 2. However, since b ≤ 1 and 2a + 2 ≥ 2*0 + 2 = 2. But 2 > 1, so there are no solutions for the second inequality. Because b cannot be ≥ 2 when b ≤ 1. Therefore, the second inequality is impossible. Therefore, the only condition is 2a + b ≥ 2.

Wait, that seems important. Let me check again.

Second inequality: 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. But since b is between 0 and 1, there is no (a, b) in [0,1]x[0,1] such that b ≥ 2a + 2. Therefore, the second inequality is never satisfied. 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.

Therefore, the area of S is the area of the region in the unit square where 2a + b ≥ 2.

So, now, I need to compute the area of the region defined by 2a + b ≥ 2 with 0 ≤ a ≤ 1, 0 ≤ b ≤ 1.

Let me sketch this mentally. The line 2a + b = 2 in the unit square. Let me find the intercepts.

When a = 0: 0 + b = 2 ⇒ b = 2, which 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 also when b = 1: 2a + 1 = 2 ⇒ 2a = 1 ⇒ a = 0.5. Therefore, the line passes through (0.5, 1).

Therefore, the line connects (1, 0) to (0.5, 1). Therefore, the region 2a + b ≥ 2 is the area above this line within the unit square.

Therefore, the area we need 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 where 2a + b ≥ 2. Since the unit square has area 1, and the region is a polygon, we can compute it by integrating or by geometric formulas.

Let me visualize the unit square with vertices at (0,0), (1,0), (1,1), (0,1). The line from (1, 0) to (0.5, 1) divides the square into two regions. The area above the line is a triangle?

Wait, let me check. The line goes from (1, 0) to (0.5, 1). So, if we consider the area above this line within the square, it's a triangle with vertices at (0.5, 1), (1, 1), and (1, 0). Wait, is that correct?

Wait, when a = 1, b can be from 0 to 1. The line at a=1 is at b=0. When a decreases from 1 to 0.5, the line goes up to b=1. So the region 2a + b ≥ 2 is above the line. So, for a between 0.5 and 1, the line goes from (0.5, 1) to (1, 0). So the area above the line would be the area bounded by the line, the top edge of the square (b=1), and the right edge (a=1). So, this is a triangle with vertices at (0.5, 1), (1, 1), and (1, 0). Let me confirm.

Yes, because for a from 0.5 to 1, the line 2a + b = 2 goes from (0.5, 1) to (1, 0). The area above the line would be the area between the line, the top of the square (b=1), and the right side (a=1). So that's a triangle with base along the right edge from (1, 0) to (1, 1), which is length 1, and the base along the top edge from (1, 1) to (0.5, 1), which is length 0.5. Wait, but actually, the triangle is formed by the points (1, 0), (1, 1), and (0.5, 1). The base is from (1, 0) to (1, 1), which is length 1, and 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 triangle has vertices at (1, 0), (1, 1), and (0.5, 1). So, plotting these points, the triangle is 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 that seems small. Wait, let me check again.

Alternatively, the area can be calculated by integrating. For a from 0.5 to 1, the lower bound for b is the line 2a + b = 2 ⇒ b = 2 - 2a. The upper bound for b is 1. Therefore, the area is the integral from a = 0.5 to 1 of [1 - (2 - 2a)] da = integral from 0.5 to 1 of (1 - 2 + 2a) da = integral from 0.5 to 1 of (2a - 1) da.

Calculating that:

Integral of 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. So the area is 0.25. Therefore, the area of the region where 2a + b ≥ 2 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". But according to our analysis, this is equivalent to 2a + b ≥ 2, which has area 1/4. However, let me check if there are other regions where the equation might have real roots. Wait, earlier we thought that the second inequality (b ≥ 2a + 2) is impossible, so only the first inequality 2a + b ≥ 2 is relevant. But let me check again.

Wait, the original condition was that the quadratic equation in y has at least one root in (-∞, -2] ∪ [2, ∞). We translated that into f(-2) ≤ 0 or f(2) ≤ 0. But we saw that f(2) ≤ 0 leads to b ≥ 2a + 2, which is impossible. However, is there a case where even though f(-2) > 0 and f(2) > 0, the quadratic equation still has a root in (-∞, -2] ∪ [2, ∞)? For example, if the quadratic has one root in (-∞, -2] and another in [2, ∞), but then f(-2) and f(2) would both be negative? Wait, no. If the quadratic has one root in (-∞, -2] and another in [2, ∞), then since the parabola opens upwards, the function is negative between the two roots. Therefore, f(-2) would be negative (since -2 is between the two roots) and f(2) would also be negative (since 2 is between the two roots). Therefore, in that case, both f(-2) ≤ 0 and 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. However, since f(2) ≤ 0 is impossible, the only condition is f(-2) ≤ 0, which is 2 - 2a - b ≤ 0 ⇒ 2a + b ≥ 2. Therefore, the area is indeed 1/4. But wait, let me check with an example.

Take a = 1, b = 0. Then 2a + b = 2*1 + 0 = 2, which is equal to 2. So this point is on the boundary. Let's check if the original equation has a real root. The equation becomes x⁴ + 1*x³ - 0*x² + 1*x + 1 = x⁴ + x³ + x + 1. Let me try to factor this. Maybe factor by grouping:

x⁴ + x³ + x + 1 = x³(x + 1) + 1(x + 1) = (x³ + 1)(x + 1). Then, x³ + 1 factors as (x + 1)(x² - x + 1). Therefore, the equation factors as (x + 1)^2 (x² - x + 1) = 0. The roots are x = -1 (double root) and roots of x² - x + 1 = 0, which are complex. Therefore, x = -1 is a real root. Therefore, the equation does have a real root. So (1, 0) is in S.

Another example: take a = 0.5, b = 1. Then 2a + b = 1 + 1 = 2. So again on the boundary. Let's check the equation: x⁴ + 0.5x³ - 1x² + 0.5x + 1 = 0. Let me try to use the substitution y = x + 1/x. Let me compute y² + a y - (b + 2) = y² + 0.5 y - (1 + 2) = y² + 0.5 y - 3 = 0. Solving for y: y = [-0.5 ± √(0.25 + 12)] / 2 = [-0.5 ± √(12.25)] / 2 = [-0.5 ± 3.5]/2. Therefore, y = (3)/2 = 1.5 or y = (-4)/2 = -2. So y = -2 is a root. Then, since y = x + 1/x = -2. Let me solve x + 1/x = -2. Multiply by x: x² + 1 = -2x ⇒ x² + 2x + 1 = 0 ⇒ (x + 1)^2 = 0 ⇒ x = -1. Therefore, x = -1 is a root. Therefore, the equation has a real root. So (0.5, 1) is in S.

Another example: take a point inside the region, say a = 1, b = 1. Then 2a + b = 2 + 1 = 3 ≥ 2. Let's check the equation: x⁴ + 1x³ - 1x² + 1x + 1 = x⁴ + x³ - x² + x + 1. Let me try to see if it has real roots. Let me check x = -1: (-1)^4 + (-1)^3 - (-1)^2 + (-1) + 1 = 1 - 1 - 1 - 1 + 1 = -1 ≠ 0. How about x = 1: 1 + 1 - 1 + 1 + 1 = 3 ≠ 0. Maybe try x = -2: 16 - 8 - 4 - 2 + 1 = 3 ≠ 0. Maybe try to use substitution again. Let me use y = x + 1/x. Then the equation becomes y² + a y - (b + 2) = y² + 1*y - (1 + 2) = y² + y - 3 = 0. Solutions: y = [-1 ± √(1 + 12)] / 2 = [-1 ± √13]/2. √13 ≈ 3.606, so y ≈ (-1 + 3.606)/2 ≈ 1.303, and y ≈ (-1 - 3.606)/2 ≈ -2.303. Now, check if these y's can be expressed as x + 1/x. For y ≈ -2.303, which is less than -2, so there exists real x such that x + 1/x = y. Therefore, the equation has real roots. Therefore, (1, 1) is in S.

Now, take a point not in the region, say a = 0, b = 0. Then 2a + b = 0 < 2. Let's check if the equation has real roots. The equation is x⁴ + 0x³ - 0x² + 0x + 1 = x⁴ + 1 = 0. Which has no real roots. So (0, 0) is not in S. Another point, say a = 0.5, b = 0. Then 2a + b = 1 + 0 = 1 < 2. Let's check the equation: x⁴ + 0.5x³ - 0x² + 0.5x + 1 = x⁴ + 0.5x³ + 0.5x + 1. Let me check for real roots. Try x = -1: 1 - 0.5 - 0.5 + 1 = 1 - 0.5 - 0.5 + 1 = 1. Not zero. Try x = -2: 16 - 4 - 1 + 1 = 12 ≠ 0. Maybe try substitution again. y = x + 1/x. Then the equation becomes y² + 0.5 y - (0 + 2) = y² + 0.5 y - 2 = 0. Solutions: y = [-0.5 ± √(0.25 + 8)] / 2 = [-0.5 ± √8.25]/2 ≈ [-0.5 ± 2.872]/2. So y ≈ (2.372)/2 ≈ 1.186 or y ≈ (-3.372)/2 ≈ -1.686. Both y ≈ 1.186 and y ≈ -1.686. However, the range of y is (-∞, -2] ∪ [2, ∞). So neither of these y's are in that range. Therefore, there are no real roots. Therefore, (0.5, 0) is not in S. Therefore, our condition seems correct.

Therefore, the set S is the region 2a + b ≥ 2 in the unit square, which is a triangle with area 1/4. Therefore, the answer is 1/4? Wait, but let me check again. Wait, the area we calculated was 0.25, which is 1/4. However, let me think again. The region 2a + b ≥ 2 in the unit square is a triangle with vertices at (1, 0), (1, 1), and (0.5, 1). The area of this triangle is indeed 0.25. However, let me check with another approach.

Alternatively, the line 2a + b = 2 intersects the square at (1, 0) and (0.5, 1). The area above this line is a triangle. The base of the triangle is from (1, 0) to (1, 1), which is length 1. The height is the horizontal distance from (1, 0) to (0.5, 1), but since it's a triangle, maybe it's better to use coordinates.

Alternatively, using coordinates, the three vertices of the triangle are (1, 0), (1, 1), and (0.5, 1). The area can be calculated using the formula for the area of a triangle with coordinates. Using the formula:

Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|

Plugging in (x1, y1) = (1, 0), (x2, y2) = (1, 1), (x3, y3) = (0.5, 1):

Area = |1*(1 - 1) + 1*(1 - 0) + 0.5*(0 - 1)| / 2

= |0 + 1*(1) + 0.5*(-1)| / 2

= |0 + 1 - 0.5| / 2

= |0.5| / 2 = 0.25.

Yes, so the area is 1/4. Therefore, the area of S is 1/4. However, 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". But according to our analysis, this is exactly the region 2a + b ≥ 2, which has area 1/4. However, I need to make sure that there are no other regions where the equation could have real roots. For example, could there be cases where the quadratic equation in y has roots in (-∞, -2] ∪ [2, ∞) even if 2a + b < 2? But according to our previous reasoning, since f(-2) = 2 - 2a - b. If 2a + b < 2, then f(-2) = 2 - 2a - b > 0. Similarly, f(2) = 2 + 2a - b. If 2a + b < 2, then 2 + 2a - b = 2 + (2a - b). But since 2a - b can be anything. Wait, but if 2a + b < 2, then 2a - b can be... Let me think. For example, take a = 0.5, b = 0. Then 2a + b = 1 < 2. Then f(2) = 2 + 2*(0.5) - 0 = 2 + 1 = 3 > 0. So f(2) > 0. Therefore, in this case, both f(-2) and f(2) are positive, so the quadratic equation has no roots in (-∞, -2] ∪ [2, ∞). Therefore, the equation has no real roots. Therefore, our conclusion that only 2a + b ≥ 2 is correct.

But let me check another case where maybe f(-2) > 0 and f(2) < 0. Wait, is that possible? For example, suppose that 2a + b < 2 and 2 + 2a - b < 0. Let me see. 2 + 2a - b < 0 ⇒ b > 2 + 2a. But since b ≤ 1 and 2 + 2a ≥ 2, this is impossible. Therefore, there is no (a, b) in [0,1]x[0,1] where f(2) < 0. Therefore, the only possible case is when f(-2) ≤ 0, which is 2a + b ≥ 2. Therefore, the area is indeed 1/4.

But wait, the problem says "the equation has at least one real root". However, when we did the substitution, we assumed that x ≠ 0, but we checked that x=0 is not a root. Therefore, all real roots are accounted for by the y substitution. Therefore, our analysis is correct.

But wait, let me check another case where maybe the quadratic equation in y has a double root in (-∞, -2] ∪ [2, ∞). For example, suppose that the quadratic equation has a double root at y = 2. Then, discriminant is zero, and y = [-a]/2. So, if y = 2 is a double root, then:

From quadratic equation y² + a y - (b + 2) = 0, if y = 2 is a double root, then:

The quadratic can be written as (y - 2)^2 = y² - 4y + 4. Comparing coefficients:

y² + a y - (b + 2) = y² - 4y + 4 ⇒ a = -4 and - (b + 2) = 4 ⇒ b = -6. But since b is between 0 and 1, this is impossible. Similarly, for y = -2, the double root would require:

(y + 2)^2 = y² + 4y + 4. Comparing with y² + a y - (b + 2) = 0:

a = 4 and - (b + 2) = 4 ⇒ b = -6. Again impossible. Therefore, there are no double roots in the required regions for (a, b) in [0,1]x[0,1]. Therefore, the only case is when the quadratic has two distinct roots, one in (-∞, -2] and one in [2, ∞), or one root in (-∞, -2] or one in [2, ∞). But since the quadratic has two roots, and the parabola opens upwards, if one root is in (-∞, -2], then the other root is in [something]. But since f(-2) ≤ 0, then the other root is ≥ -2. However, for the equation to have a real root, we only need at least one root in (-∞, -2] ∪ [2, ∞). Therefore, our previous conclusion holds.

Therefore, the area of S is 1/4. However, wait, I have a doubt here. Because when I calculated the area, I found 1/4, but maybe I need to check if there are other regions where the equation could have real roots. For example, suppose that the quadratic equation in y has a root in (-∞, -2], but we have already considered that by the condition 2a + b ≥ 2. But maybe there are other cases?

Wait, let me think again. The original equation is quartic. Maybe there are cases where even if the quadratic in y doesn't have roots in (-∞, -2] ∪ [2, ∞), the original equation could have real roots? For example, if the substitution y = x + 1/x is not valid? But we already checked that x=0 is not a root, so all real roots satisfy x ≠ 0, so substitution is valid. Therefore, the original equation has real roots if and only if the quadratic in y has at least one root in (-∞, -2] ∪ [2, ∞). Therefore, our analysis is correct.

Therefore, the area is 1/4. However, I recall that sometimes in these problems, the answer might be different. Let me check again with another approach.

Alternative approach: Let me consider the original equation x⁴ + a x³ - b x² + a x + 1 = 0. Let me try to factor it. Since it's palindromic, as we saw earlier, we can write it as x² + a x - b + a / x + 1 / x² = 0, which becomes y² + a y - (b + 2) = 0 where y = x + 1/x. Therefore, the equation has real roots iff this quadratic has real solutions y with |y| ≥ 2. Therefore, the condition is that the quadratic equation has at least one real root with |y| ≥ 2. Which we translated into f(-2) ≤ 0 or f(2) ≤ 0. Which led us to 2a + b ≥ 2. Therefore, the area is 1/4.

But let me check with another example. Suppose a = 0.6, b = 0.8. Then 2a + b = 1.2 + 0.8 = 2.0. So on the boundary. Let me check the equation. The equation becomes x⁴ + 0.6x³ - 0.8x² + 0.6x + 1 = 0. Let me use substitution y = x + 1/x. Then the quadratic equation is y² + 0.6 y - (0.8 + 2) = y² + 0.6 y - 2.8 = 0. Solving: y = [-0.6 ± √(0.36 + 11.2)] / 2 = [-0.6 ± √11.56]/2. √11.56 = 3.4. Therefore, y = (-0.6 + 3.4)/2 = 2.8 / 2 = 1.4 or y = (-0.6 - 3.4)/2 = -4 / 2 = -2. So y = -2 is a root. Then, solving x + 1/x = -2 gives x = -1 as before. Therefore, the equation has a real root. Therefore, (0.6, 0.8) is in S.

Another example: take a = 0.75, b = 0.5. Then 2a + b = 1.5 + 0.5 = 2.0. So on the boundary. The quadratic equation is y² + 0.75 y - (0.5 + 2) = y² + 0.75 y - 2.5 = 0. Solutions: y = [-0.75 ± √(0.5625 + 10)] / 2 = [-0.75 ± √10.5625]/2. √10.5625 = 3.25. Therefore, y = (-0.75 + 3.25)/2 = 2.5 / 2 = 1.25 or y = (-0.75 - 3.25)/2 = -4 / 2 = -2. Again, y = -2 is a root, leading to x = -1. Therefore, real root exists.

Therefore, the boundary is correct. Therefore, the area is 1/4. However, I have a lingering doubt because sometimes these problems have more complex regions. Wait, but according to the analysis, the only region is 2a + b ≥ 2, which is a triangle with area 1/4. However, let me check the total area of the unit square is 1, and 1/4 seems small, but given that the line cuts off a triangle, it might be correct.

Alternatively, maybe I made a mistake in translating the condition. Let me check again.

Original condition: the quadratic equation in y has at least one root in (-∞, -2] ∪ [2, ∞). Which we translated into f(-2) ≤ 0 or f(2) ≤ 0. But is that correct?

Suppose that the quadratic equation has a root in [2, ∞). Then, since the parabola opens upwards, f(2) ≤ 0. Similarly, if it has a root in (-∞, -2], then f(-2) ≤ 0. Therefore, the condition is f(-2) ≤ 0 or f(2) ≤ 0. However, since f(2) ≤ 0 is impossible, only f(-2) ≤ 0 is required. Therefore, the region is 2 - 2a - b ≤ 0 ⇒ 2a + b ≥ 2. Therefore, area is 1/4.

Alternatively, maybe there's a mistake in assuming that f(-2) ≤ 0 is equivalent to having a root ≤ -2. Let me think. Suppose f(-2) ≤ 0. Since the parabola opens upwards, then there are two real roots. The smaller root is ≤ -2 and the larger root is ≥ -2. Therefore, there is a root ≤ -2. Therefore, the original equation has a real root. Conversely, if the original equation has a real root, then there exists y with |y| ≥ 2, so either y ≥ 2 or y ≤ -2. If y ≥ 2, then since the parabola opens upwards, f(2) ≤ 0. If y ≤ -2, then f(-2) ≤ 0. Therefore, the condition is f(-2) ≤ 0 or f(2) ≤ 0. Therefore, the analysis is correct.

Therefore, the area is 1/4. However, I recall that sometimes when you have palindromic polynomials, there might be other roots or something. Wait, but we already considered all real roots via substitution. Therefore, I think the answer is 1/4.

But let me check with another method. Let me consider the original equation x⁴ + a x³ - b x² + a x + 1 = 0. Let me try to write it as x⁴ + 1 + a x³ + a x - b x² = 0. Note that x⁴ + 1 can be written as (x² + 1)^2 - 2x². Therefore, the equation becomes (x² + 1)^2 - 2x² + a x (x² + 1) - b x² = 0. Not sure if helpful.

Alternatively, maybe factor the equation. Let me try to factor it as (x² + p x + 1)(x² + q x + 1) = 0. Let me check if this is possible. Multiplying out:

(x² + p x + 1)(x² + q x + 1) = x⁴ + (p + q)x³ + (pq + 2)x² + (p + q)x + 1.

Comparing with original equation x⁴ + a x³ - b x² + a x + 1. Therefore, we have:

p + q = a,

pq + 2 = -b,

p + q = a.

Therefore, from the first and third equations, p + q = a. From the second equation, pq = -b - 2.

Therefore, if we can find real numbers p and q such that p + q = a and pq = -b - 2, then the original equation factors into two quadratics, each of which may have real roots. However, even if it doesn't factor, the existence of real roots can be determined by the discriminant of the quadratics. However, this seems more complicated. However, if we suppose that the original equation has real roots, then there exists real x such that the equation holds. If we can write it as product of two quadratics, then each quadratic may have real roots. However, this approach might not be necessary.

Alternatively, since we already have a solution via substitution, and verified with examples, maybe it's safe to go with 1/4. However, I have a doubt because sometimes problems like this have areas that are more complex. Wait, let me check the total area again. The region 2a + b ≥ 2 in the unit square is a triangle with vertices at (1, 0), (1, 1), and (0.5, 1). The area is 1/4. However, if I imagine the unit square, the line from (1, 0) to (0.5, 1) divides the square into a triangle of area 1/4 and a pentagon of area 3/4. But maybe I need to check with integration.

Alternatively, think of the region 2a + b ≥ 2. For a between 0.5 and 1, b ranges from 2 - 2a to 1. 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, which we calculated as 0.25. Therefore, yes, 1/4.

But let me check with another perspective. The line 2a + b = 2 is equivalent to b = 2 - 2a. When a = 0.5, b = 1. When a = 1, b = 0. So the area above this line is the set of points where b ≥ 2 - 2a. For a from 0.5 to 1, and for a from 0 to 0.5, 2 - 2a ≥ 2 - 2*(0.5) = 1. But since b ≤ 1, for a < 0.5, 2 - 2a > 1, so the condition b ≥ 2 - 2a would require b ≥ something greater than 1, which is impossible. Therefore, the region is only for a from 0.5 to 1, and b from 2 - 2a to 1. Therefore, the area is indeed the integral from 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, 1/4.

Therefore, I think the answer is 1/4. However, I recall that in some similar problems, the area might be different. Wait, but let me check with another example. Suppose a = 0.75, b = 0.5. Then 2a + b = 1.5 + 0.5 = 2. So on the boundary. The equation has a real root. If I take a = 0.75, b = 0. Then 2a + b = 1.5 < 2. The equation is x⁴ + 0.75x³ - 0x² + 0.75x + 1 = x⁴ + 0.75x³ + 0.75x + 1. Let me check for real roots. Try x = -1: 1 - 0.75 - 0.75 + 1 = 1 - 0.75 - 0.75 + 1 = 0.5 ≠ 0. Try x = -2: 16 - 12 - 1.5 + 1 = 3.5 ≠ 0. Try x = -0.5: (0.0625) + 0.75*(-0.125) + 0.75*(-0.5) + 1 ≈ 0.0625 - 0.09375 - 0.375 + 1 ≈ 0.6. Not zero. Therefore, no real roots. Therefore, the area is indeed 1/4.

Therefore, I think the answer is 1/4, so the area of S is 1/4. Therefore, the 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 some $ (a, b) $ with $ 0 \le a, b \le 1 $. Our goal is to find the area of the region in the unit square $ [0,1] \times [0,1] $ where this condition is satisfied.

---

### Step 1: Recognize the Polynomial Structure

The polynomial is **palindromic**, meaning its coefficients are symmetric. That is, the coefficients of $ x^4 $ and the constant term are both 1, and the coefficients of $ x^3 $ and $ x $ are both $ a $. This symmetry allows us to use a substitution to simplify the equation.

Let us divide the entire equation by $ x^2 $ (valid for $ x \ne 0 $, which we will verify later):

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

Now, define $ y = x + \frac{1}{x} $. Then, we can use the identity:

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

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 **real solutions** $ 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 $ y = x + \frac{1}{x} $ has the following range:

- If $ x > 0 $, then $ y \ge 2 $ by AM-GM inequality.
- If $ x < 0 $, then $ y \le -2 $.

So, the possible values of $ y $ are in the union of intervals:

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

Thus, the original equation has a real root if and only if the quadratic in $ y $ has **at least one real root** in this union.

---

### 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), this happens if and only if **at least one of the following is true**:

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

Let’s compute both:

- $ 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 condition becomes:

$$
2 - 2a - b \le 0 \quad \text{or} \quad 2 + 2a - b \le 0
$$

Simplify both:

- $ 2a + b \ge 2 $
- $ 2a - b \le -2 \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 **triangle** bounded by the line $ 2a + b = 2 $, the vertical line $ a = 1 $, and the horizontal line $ b = 1 $.

- The line $ 2a + b = 2 $ intersects the square at:
  - $ a = 1 \Rightarrow b = 0 $
  - $ b = 1 \Rightarrow a = 0.5 $

So the triangle has vertices at:

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

This is a right triangle with base $ 1 $ (from $ (1,0) $ to $ (1,1) $) and height $ 0.5 $ (from $ (1,1) $ to $ (0.5,1) $).

The area of this triangle is:

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

---

### 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 one root in (-∞, -2] and another root in [2, ∞). Then, since the parabola opens upwards, f(-2) would be negative (since between the two roots, the function is negative). Similarly, f(2) would be negative. Therefore, if there is a root in (-∞, -2] and another in [2, ∞), then both f(-2) ≤ 0 and 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. Wait, but if f(-2) ≤ 0 or f(2) ≤ 0, then there is at least one root in the required intervals. 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 union of the sets where f(-2) ≤ 0 or f(2) ≤ 0.

Therefore, we can write:

Original equation has real root ⇨ f(-2) ≤ 0 or f(2) ≤ 0.

Therefore, let's compute f(-2) and f(2):

First, f(y) = y² + a y - (b + 2).

Compute f(-2):

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

Therefore, we need to find the area of the region in the unit square [0,1]x[0,1] where 2 - 2a - b ≤ 0 or 2 + 2a - b ≤ 0.

Let me write these inequalities:

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

Second inequality: 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 second inequality: b ≥ 2a + 2. However, since b ≤ 1 and 2a + 2 ≥ 2*0 + 2 = 2. But 2 > 1, so there are no solutions for the second inequality. Because b cannot be ≥ 2 when b ≤ 1. Therefore, the second inequality is impossible. Therefore, the only condition is 2a + b ≥ 2.

Wait, that seems important. Let me check again.

Second inequality: 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. But since b is between 0 and 1, there is no (a, b) in [0,1]x[0,1] such that b ≥ 2a + 2. Therefore, the second inequality is never satisfied. 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.

Therefore, the area of S is the area of the region in the unit square where 2a + b ≥ 2.

So, now, I need to compute the area of the region defined by 2a + b ≥ 2 with 0 ≤ a ≤ 1, 0 ≤ b ≤ 1.

Let me sketch this mentally. The line 2a + b = 2 in the unit square. Let me find the intercepts.

When a = 0: 0 + b = 2 ⇒ b = 2, which 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 also when b = 1: 2a + 1 = 2 ⇒ 2a = 1 ⇒ a = 0.5. Therefore, the line passes through (0.5, 1).

Therefore, the line connects (1, 0) to (0.5, 1). Therefore, the region 2a + b ≥ 2 is the area above this line within the unit square.

Therefore, the area we need 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 where 2a + b ≥ 2. Since the unit square has area 1, and the region is a polygon, we can compute it by integrating or by geometric formulas.

Let me visualize the unit square with vertices at (0,0), (1,0), (1,1), (0,1). The line from (1, 0) to (0.5, 1) divides the square into two regions. The area above the line is a triangle?

Wait, let me check. The line goes from (1, 0) to (0.5, 1). So, if we consider the area above this line within the square, it's a triangle with vertices at (0.5, 1), (1, 1), and (1, 0). Wait, is that correct?

Wait, when a = 1, b can be from 0 to 1. The line at a=1 is at b=0. When a decreases from 1 to 0.5, the line goes up to b=1. So the region 2a + b ≥ 2 is above the line. So, for a between 0.5 and 1, the line goes from (0.5, 1) to (1, 0). So the area above the line would be the area bounded by the line, the top edge of the square (b=1), and the right edge (a=1). So, this is a triangle with vertices at (0.5, 1), (1, 1), and (1, 0). Let me confirm.

Yes, because for a from 0.5 to 1, the line 2a + b = 2 goes from (0.5, 1) to (1, 0). The area above the line would be the area between the line, the top of the square (b=1), and the right side (a=1). So that's a triangle with base along the right edge from (1, 0) to (1, 1), which is length 1, and the base along the top edge from (1, 1) to (0.5, 1), which is length 0.5. Wait, but actually, the triangle is formed by the points (1, 0), (1, 1), and (0.5, 1). The base is from (1, 0) to (1, 1), which is length 1, and 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 triangle has vertices at (1, 0), (1, 1), and (0.5, 1). So, plotting these points, the triangle is 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 that seems small. Wait, let me check again.

Alternatively, the area can be calculated by integrating. For a from 0.5 to 1, the lower bound for b is the line 2a + b = 2 ⇒ b = 2 - 2a. The upper bound for b is 1. Therefore, the area is the integral from a = 0.5 to 1 of [1 - (2 - 2a)] da = integral from 0.5 to 1 of (1 - 2 + 2a) da = integral from 0.5 to 1 of (2a - 1) da.

Calculating that:

Integral of 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. So the area is 0.25. Therefore, the area of the region where 2a + b ≥ 2 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". But according to our analysis, this is equivalent to 2a + b ≥ 2, which has area 1/4. However, let me check if there are other regions where the equation might have real roots. Wait, earlier we thought that the second inequality (b ≥ 2a + 2) is impossible, so only the first inequality 2a + b ≥ 2 is relevant. But let me check again.

Wait, the original condition was that the quadratic equation in y has at least one root in (-∞, -2] ∪ [2, ∞). We translated that into f(-2) ≤ 0 or f(2) ≤ 0. But we saw that f(2) ≤ 0 leads to b ≥ 2a + 2, which is impossible. However, is there a case where even though f(-2) > 0 and f(2) > 0, the quadratic equation still has a root in (-∞, -2] ∪ [2, ∞)? For example, if the quadratic has one root in (-∞, -2] and another in [2, ∞), but then f(-2) and f(2) would both be negative? Wait, no. If the quadratic has one root in (-∞, -2] and another in [2, ∞), then since the parabola opens upwards, the function is negative between the two roots. Therefore, f(-2) would be negative (since -2 is between the two roots) and f(2) would also be negative (since 2 is between the two roots). Therefore, in that case, both f(-2) ≤ 0 and 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. However, since f(2) ≤ 0 is impossible, the only condition is f(-2) ≤ 0, which is 2 - 2a - b ≤ 0 ⇒ 2a + b ≥ 2. Therefore, the area is indeed 1/4. But wait, let me check with an example.

Take a = 1, b = 0. Then 2a + b = 2*1 + 0 = 2, which is equal to 2. So this point is on the boundary. Let's check if the original equation has a real root. The equation becomes x⁴ + 1*x³ - 0*x² + 1*x + 1 = x⁴ + x³ + x + 1. Let me try to factor this. Maybe factor by grouping:

x⁴ + x³ + x + 1 = x³(x + 1) + 1(x + 1) = (x³ + 1)(x + 1). Then, x³ + 1 factors as (x + 1)(x² - x + 1). Therefore, the equation factors as (x + 1)^2 (x² - x + 1) = 0. The roots are x = -1 (double root) and roots of x² - x + 1 = 0, which are complex. Therefore, x = -1 is a real root. Therefore, the equation does have a real root. So (1, 0) is in S.

Another example: take a = 0.5, b = 1. Then 2a + b = 1 + 1 = 2. So again on the boundary. Let's check the equation: x⁴ + 0.5x³ - 1x² + 0.5x + 1 = 0. Let me try to use the substitution y = x + 1/x. Let me compute y² + a y - (b + 2) = y² + 0.5 y - (1 + 2) = y² + 0.5 y - 3 = 0. Solving for y: y = [-0.5 ± √(0.25 + 12)] / 2 = [-0.5 ± √(12.25)] / 2 = [-0.5 ± 3.5]/2. Therefore, y = (3)/2 = 1.5 or y = (-4)/2 = -2. So y = -2 is a root. Then, since y = x + 1/x = -2. Let me solve x + 1/x = -2. Multiply by x: x² + 1 = -2x ⇒ x² + 2x + 1 = 0 ⇒ (x + 1)^2 = 0 ⇒ x = -1. Therefore, x = -1 is a root. Therefore, the equation has a real root. So (0.5, 1) is in S.

Another example: take a point inside the region, say a = 1, b = 1. Then 2a + b = 2 + 1 = 3 ≥ 2. Let's check the equation: x⁴ + 1x³ - 1x² + 1x + 1 = x⁴ + x³ - x² + x + 1. Let me try to see if it has real roots. Let me check x = -1: (-1)^4 + (-1)^3 - (-1)^2 + (-1) + 1 = 1 - 1 - 1 - 1 + 1 = -1 ≠ 0. How about x = 1: 1 + 1 - 1 + 1 + 1 = 3 ≠ 0. Maybe try x = -2: 16 - 8 - 4 - 2 + 1 = 3 ≠ 0. Maybe try to use substitution again. Let me use y = x + 1/x. Then the equation becomes y² + a y - (b + 2) = y² + 1*y - (1 + 2) = y² + y - 3 = 0. Solutions: y = [-1 ± √(1 + 12)] / 2 = [-1 ± √13]/2. √13 ≈ 3.606, so y ≈ (-1 + 3.606)/2 ≈ 1.303, and y ≈ (-1 - 3.606)/2 ≈ -2.303. Now, check if these y's can be expressed as x + 1/x. For y ≈ -2.303, which is less than -2, so there exists real x such that x + 1/x = y. Therefore, the equation has real roots. Therefore, (1, 1) is in S.

Now, take a point not in the region, say a = 0, b = 0. Then 2a + b = 0 < 2. Let's check if the equation has real roots. The equation is x⁴ + 0x³ - 0x² + 0x + 1 = x⁴ + 1 = 0. Which has no real roots. So (0, 0) is not in S. Another point, say a = 0.5, b = 0. Then 2a + b = 1 + 0 = 1 < 2. Let's check the equation: x⁴ + 0.5x³ - 0x² + 0.5x + 1 = x⁴ + 0.5x³ + 0.5x + 1. Let me check for real roots. Try x = -1: 1 - 0.5 - 0.5 + 1 = 1 - 0.5 - 0.5 + 1 = 1. Not zero. Try x = -2: 16 - 4 - 1 + 1 = 12 ≠ 0. Maybe try substitution again. y = x + 1/x. Then the equation becomes y² + 0.5 y - (0 + 2) = y² + 0.5 y - 2 = 0. Solutions: y = [-0.5 ± √(0.25 + 8)] / 2 = [-0.5 ± √8.25]/2 ≈ [-0.5 ± 2.872]/2. So y ≈ (2.372)/2 ≈ 1.186 or y ≈ (-3.372)/2 ≈ -1.686. Both y ≈ 1.186 and y ≈ -1.686. However, the range of y is (-∞, -2] ∪ [2, ∞). So neither of these y's are in that range. Therefore, there are no real roots. Therefore, (0.5, 0) is not in S. Therefore, our condition seems correct.

Therefore, the set S is the region 2a + b ≥ 2 in the unit square, which is a triangle with area 1/4. Therefore, the answer is 1/4? Wait, but let me check again. Wait, the area we calculated was 0.25, which is 1/4. However, let me think again. The region 2a + b ≥ 2 in the unit square is a triangle with vertices at (1, 0), (1, 1), and (0.5, 1). The area of this triangle is indeed 0.25. However, let me check with another approach.

Alternatively, the line 2a + b = 2 intersects the square at (1, 0) and (0.5, 1). The area above this line is a triangle. The base of the triangle is from (1, 0) to (1, 1), which is length 1. The height is the horizontal distance from (1, 0) to (0.5, 1), but since it's a triangle, maybe it's better to use coordinates.

Alternatively, using coordinates, the three vertices of the triangle are (1, 0), (1, 1), and (0.5, 1). The area can be calculated using the formula for the area of a triangle with coordinates. Using the formula:

Area = |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))/2|

Plugging in (x1, y1) = (1, 0), (x2, y2) = (1, 1), (x3, y3) = (0.5, 1):

Area = |1*(1 - 1) + 1*(1 - 0) + 0.5*(0 - 1)| / 2

= |0 + 1*(1) + 0.5*(-1)| / 2

= |0 + 1 - 0.5| / 2

= |0.5| / 2 = 0.25.

Yes, so the area is 1/4. Therefore, the area of S is 1/4. However, 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". But according to our analysis, this is exactly the region 2a + b ≥ 2, which has area 1/4. However, I need to make sure that there are no other regions where the equation could have real roots. For example, could there be cases where the quadratic equation in y has roots in (-∞, -2] ∪ [2, ∞) even if 2a + b < 2? But according to our previous reasoning, since f(-2) = 2 - 2a - b. If 2a + b < 2, then f(-2) = 2 - 2a - b > 0. Similarly, f(2) = 2 + 2a - b. If 2a + b < 2, then 2 + 2a - b = 2 + (2a - b). But since 2a - b can be anything. Wait, but if 2a + b < 2, then 2a - b can be... Let me think. For example, take a = 0.5, b = 0. Then 2a + b = 1 < 2. Then f(2) = 2 + 2*(0.5) - 0 = 2 + 1 = 3 > 0. So f(2) > 0. Therefore, in this case, both f(-2) and f(2) are positive, so the quadratic equation has no roots in (-∞, -2] ∪ [2, ∞). Therefore, the equation has no real roots. Therefore, our conclusion that only 2a + b ≥ 2 is correct.

But let me check another case where maybe f(-2) > 0 and f(2) < 0. Wait, is that possible? For example, suppose that 2a + b < 2 and 2 + 2a - b < 0. Let me see. 2 + 2a - b < 0 ⇒ b > 2 + 2a. But since b ≤ 1 and 2 + 2a ≥ 2, this is impossible. Therefore, there is no (a, b) in [0,1]x[0,1] where f(2) < 0. Therefore, the only possible case is when f(-2) ≤ 0, which is 2a + b ≥ 2. Therefore, the area is indeed 1/4.

But wait, the problem says "the equation has at least one real root". However, when we did the substitution, we assumed that x ≠ 0, but we checked that x=0 is not a root. Therefore, all real roots are accounted for by the y substitution. Therefore, our analysis is correct.

But wait, let me check another case where maybe the quadratic equation in y has a double root in (-∞, -2] ∪ [2, ∞). For example, suppose that the quadratic equation has a double root at y = 2. Then, discriminant is zero, and y = [-a]/2. So, if y = 2 is a double root, then:

From quadratic equation y² + a y - (b + 2) = 0, if y = 2 is a double root, then:

The quadratic can be written as (y - 2)^2 = y² - 4y + 4. Comparing coefficients:

y² + a y - (b + 2) = y² - 4y + 4 ⇒ a = -4 and - (b + 2) = 4 ⇒ b = -6. But since b is between 0 and 1, this is impossible. Similarly, for y = -2, the double root would require:

(y + 2)^2 = y² + 4y + 4. Comparing with y² + a y - (b + 2) = 0:

a = 4 and - (b + 2) = 4 ⇒ b = -6. Again impossible. Therefore, there are no double roots in the required regions for (a, b) in [0,1]x[0,1]. Therefore, the only case is when the quadratic has two distinct roots, one in (-∞, -2] and one in [2, ∞), or one root in (-∞, -2] or one in [2, ∞). But since the quadratic has two roots, and the parabola opens upwards, if one root is in (-∞, -2], then the other root is in [something]. But since f(-2) ≤ 0, then the other root is ≥ -2. However, for the equation to have a real root, we only need at least one root in (-∞, -2] ∪ [2, ∞). Therefore, our previous conclusion holds.

Therefore, the area of S is 1/4. However, wait, I have a doubt here. Because when I calculated the area, I found 1/4, but maybe I need to check if there are other regions where the equation could have real roots. For example, suppose that the quadratic equation in y has a root in (-∞, -2], but we have already considered that by the condition 2a + b ≥ 2. But maybe there are other cases?

Wait, let me think again. The original equation is quartic. Maybe there are cases where even if the quadratic in y doesn't have roots in (-∞, -2] ∪ [2, ∞), the original equation could have real roots? For example, if the substitution y = x + 1/x is not valid? But we already checked that x=0 is not a root, so all real roots satisfy x ≠ 0, so substitution is valid. Therefore, the original equation has real roots if and only if the quadratic in y has at least one root in (-∞, -2] ∪ [2, ∞). Therefore, our analysis is correct.

Therefore, the area is 1/4. However, I recall that sometimes in these problems, the answer might be different. Let me check again with another approach.

Alternative approach: Let me consider the original equation x⁴ + a x³ - b x² + a x + 1 = 0. Let me try to factor it. Since it's palindromic, as we saw earlier, we can write it as x² + a x - b + a / x + 1 / x² = 0, which becomes y² + a y - (b + 2) = 0 where y = x + 1/x. Therefore, the equation has real roots iff this quadratic has real solutions y with |y| ≥ 2. Therefore, the condition is that the quadratic equation has at least one real root with |y| ≥ 2. Which we translated into f(-2) ≤ 0 or f(2) ≤ 0. Which led us to 2a + b ≥ 2. Therefore, the area is 1/4.

But let me check with another example. Suppose a = 0.6, b = 0.8. Then 2a + b = 1.2 + 0.8 = 2.0. So on the boundary. Let me check the equation. The equation becomes x⁴ + 0.6x³ - 0.8x² + 0.6x + 1 = 0. Let me use substitution y = x + 1/x. Then the quadratic equation is y² + 0.6 y - (0.8 + 2) = y² + 0.6 y - 2.8 = 0. Solving: y = [-0.6 ± √(0.36 + 11.2)] / 2 = [-0.6 ± √11.56]/2. √11.56 = 3.4. Therefore, y = (-0.6 + 3.4)/2 = 2.8 / 2 = 1.4 or y = (-0.6 - 3.4)/2 = -4 / 2 = -2. So y = -2 is a root. Then, solving x + 1/x = -2 gives x = -1 as before. Therefore, the equation has a real root. Therefore, (0.6, 0.8) is in S.

Another example: take a = 0.75, b = 0.5. Then 2a + b = 1.5 + 0.5 = 2.0. So on the boundary. The quadratic equation is y² + 0.75 y - (0.5 + 2) = y² + 0.75 y - 2.5 = 0. Solutions: y = [-0.75 ± √(0.5625 + 10)] / 2 = [-0.75 ± √10.5625]/2. √10.5625 = 3.25. Therefore, y = (-0.75 + 3.25)/2 = 2.5 / 2 = 1.25 or y = (-0.75 - 3.25)/2 = -4 / 2 = -2. Again, y = -2 is a root, leading to x = -1. Therefore, real root exists.

Therefore, the boundary is correct. Therefore, the area is 1/4. However, I have a lingering doubt because sometimes these problems have more complex regions. Wait, but according to the analysis, the only region is 2a + b ≥ 2, which is a triangle with area 1/4. However, let me check the total area of the unit square is 1, and 1/4 seems small, but given that the line cuts off a triangle, it might be correct.

Alternatively, maybe I made a mistake in translating the condition. Let me check again.

Original condition: the quadratic equation in y has at least one root in (-∞, -2] ∪ [2, ∞). Which we translated into f(-2) ≤ 0 or f(2) ≤ 0. But is that correct?

Suppose that the quadratic equation has a root in [2, ∞). Then, since the parabola opens upwards, f(2) ≤ 0. Similarly, if it has a root in (-∞, -2], then f(-2) ≤ 0. Therefore, the condition is f(-2) ≤ 0 or f(2) ≤ 0. However, since f(2) ≤ 0 is impossible, only f(-2) ≤ 0 is required. Therefore, the region is 2 - 2a - b ≤ 0 ⇒ 2a + b ≥ 2. Therefore, area is 1/4.

Alternatively, maybe there's a mistake in assuming that f(-2) ≤ 0 is equivalent to having a root ≤ -2. Let me think. Suppose f(-2) ≤ 0. Since the parabola opens upwards, then there are two real roots. The smaller root is ≤ -2 and the larger root is ≥ -2. Therefore, there is a root ≤ -2. Therefore, the original equation has a real root. Conversely, if the original equation has a real root, then there exists y with |y| ≥ 2, so either y ≥ 2 or y ≤ -2. If y ≥ 2, then since the parabola opens upwards, f(2) ≤ 0. If y ≤ -2, then f(-2) ≤ 0. Therefore, the condition is f(-2) ≤ 0 or f(2) ≤ 0. Therefore, the analysis is correct.

Therefore, the area is 1/4. However, I recall that sometimes when you have palindromic polynomials, there might be other roots or something. Wait, but we already considered all real roots via substitution. Therefore, I think the answer is 1/4.

But let me check with another method. Let me consider the original equation x⁴ + a x³ - b x² + a x + 1 = 0. Let me try to write it as x⁴ + 1 + a x³ + a x - b x² = 0. Note that x⁴ + 1 can be written as (x² + 1)^2 - 2x². Therefore, the equation becomes (x² + 1)^2 - 2x² + a x (x² + 1) - b x² = 0. Not sure if helpful.

Alternatively, maybe factor the equation. Let me try to factor it as (x² + p x + 1)(x² + q x + 1) = 0. Let me check if this is possible. Multiplying out:

(x² + p x + 1)(x² + q x + 1) = x⁴ + (p + q)x³ + (pq + 2)x² + (p + q)x + 1.

Comparing with original equation x⁴ + a x³ - b x² + a x + 1. Therefore, we have:

p + q = a,

pq + 2 = -b,

p + q = a.

Therefore, from the first and third equations, p + q = a. From the second equation, pq = -b - 2.

Therefore, if we can find real numbers p and q such that p + q = a and pq = -b - 2, then the original equation factors into two quadratics, each of which may have real roots. However, even if it doesn't factor, the existence of real roots can be determined by the discriminant of the quadratics. However, this seems more complicated. However, if we suppose that the original equation has real roots, then there exists real x such that the equation holds. If we can write it as product of two quadratics, then each quadratic may have real roots. However, this approach might not be necessary.

Alternatively, since we already have a solution via substitution, and verified with examples, maybe it's safe to go with 1/4. However, I have a doubt because sometimes problems like this have areas that are more complex. Wait, let me check the total area again. The region 2a + b ≥ 2 in the unit square is a triangle with vertices at (1, 0), (1, 1), and (0.5, 1). The area is 1/4. However, if I imagine the unit square, the line from (1, 0) to (0.5, 1) divides the square into a triangle of area 1/4 and a pentagon of area 3/4. But maybe I need to check with integration.

Alternatively, think of the region 2a + b ≥ 2. For a between 0.5 and 1, b ranges from 2 - 2a to 1. 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, which we calculated as 0.25. Therefore, yes, 1/4.

But let me check with another perspective. The line 2a + b = 2 is equivalent to b = 2 - 2a. When a = 0.5, b = 1. When a = 1, b = 0. So the area above this line is the set of points where b ≥ 2 - 2a. For a from 0.5 to 1, and for a from 0 to 0.5, 2 - 2a ≥ 2 - 2*(0.5) = 1. But since b ≤ 1, for a < 0.5, 2 - 2a > 1, so the condition b ≥ 2 - 2a would require b ≥ something greater than 1, which is impossible. Therefore, the region is only for a from 0.5 to 1, and b from 2 - 2a to 1. Therefore, the area is indeed the integral from 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, 1/4.

Therefore, I think the answer is 1/4. However, I recall that in some similar problems, the area might be different. Wait, but let me check with another example. Suppose a = 0.75, b = 0.5. Then 2a + b = 1.5 + 0.5 = 2. So on the boundary. The equation has a real root. If I take a = 0.75, b = 0. Then 2a + b = 1.5 < 2. The equation is x⁴ + 0.75x³ - 0x² + 0.75x + 1 = x⁴ + 0.75x³ + 0.75x + 1. Let me check for real roots. Try x = -1: 1 - 0.75 - 0.75 + 1 = 1 - 0.75 - 0.75 + 1 = 0.5 ≠ 0. Try x = -2: 16 - 12 - 1.5 + 1 = 3.5 ≠ 0. Try x = -0.5: (0.0625) + 0.75*(-0.125) + 0.75*(-0.5) + 1 ≈ 0.0625 - 0.09375 - 0.375 + 1 ≈ 0.6. Not zero. Therefore, no real roots. Therefore, the area is indeed 1/4.

Therefore, I think the answer is 1/4, so the area of S is 1/4. Therefore, the 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 some $ (a, b) $ with $ 0 \le a, b \le 1 $. Our goal is to find the area of the region in the unit square $ [0,1] \times [0,1] $ where this condition is satisfied.

---

### Step 1: Recognize the Polynomial Structure

The polynomial is **palindromic**, meaning its coefficients are symmetric. That is, the coefficients of $ x^4 $ and the constant term are both 1, and the coefficients of $ x^3 $ and $ x $ are both $ a $. This symmetry allows us to use a substitution to simplify the equation.

Let us divide the entire equation by $ x^2 $ (valid for $ x \ne 0 $, which we will verify later):

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

Now, define $ y = x + \frac{1}{x} $. Then, we can use the identity:

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

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 **real solutions** $ 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 $ y = x + \frac{1}{x} $ has the following range:

- If $ x > 0 $, then $ y \ge 2 $ by AM-GM inequality.
- If $ x < 0 $, then $ y \le -2 $.

So, the possible values of $ y $ are in the union of intervals:

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

Thus, the original equation has a real root if and only if the quadratic in $ y $ has **at least one real root** in this union.

---

### 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), this happens if and only if **at least one of the following is true**:

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

Let’s compute both:

- $ 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 condition becomes:

$$
2 - 2a - b \le 0 \quad \text{or} \quad 2 + 2a - b \le 0
$$

Simplify both:

- $ 2a + b \ge 2 $
- $ 2a - b \le -2 \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 **triangle** bounded by the line $ 2a + b = 2 $, the vertical line $ a = 1 $, and the horizontal line $ b = 1 $.

- The line $ 2a + b = 2 $ intersects the square at:
  - $ a = 1 \Rightarrow b = 0 $
  - $ b = 1 \Rightarrow a = 0.5 $

So the triangle has vertices at:

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

This is a right triangle with base $ 1 $ (from $ (1,0) $ to $ (1,1) $) and height $ 0.5 $ (from $ (1,1) $ to $ (0.5,1) $).

The area of this triangle is:

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

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### 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 $\boxed{\frac{1}{4}}.$