Explanation:
Okay, so I need to find the area of the set S, which consists of all 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. Alright, let me start by understanding the problem.

First, the equation given is a quartic polynomial: x⁴ + a x³ - b x² + a x + 1 = 0. The coefficients a and b are between 0 and 1. The question is asking for the area of all (a, b) pairs where this equation has at least one real root.

So, my goal is to figure out the region in the [0,1]x[0,1] square where this polynomial has at least one real root, and then compute the area of that region.

Let me recall that for polynomials, especially quartic ones, it's not straightforward to determine the number of real roots. However, maybe there's a way to simplify this equation or find some conditions on a and b that would guarantee a real root.

Let me first check if there are any obvious substitutions or symmetries. The polynomial is x⁴ + a x³ - b x² + a x + 1. If I look at the coefficients, the coefficients of x⁴ and the constant term are both 1. Also, the coefficients of x³ and x are both a. That symmetry might be useful.

Perhaps I can factor this polynomial or rewrite it in a way that makes it easier to analyze. Let me try to see if it can be written as a quadratic in terms of x + 1/x or something like that.

Let me consider dividing both sides by x² to make it symmetric. Let me try that:

Dividing by x²:

x² + a x - b + a / x + 1 / x² = 0

Hmm, let me rewrite this:

x² + 1/x² + a(x + 1/x) - b = 0

Yes, that seems promising. Let me set y = x + 1/x. Then, note that y² = x² + 2 + 1/x², so x² + 1/x² = y² - 2.

Therefore, substituting into the equation:

(y² - 2) + a y - b = 0 ⇒ y² + a y - (b + 2) = 0

So, the original quartic equation is equivalent to the quadratic equation in y: y² + a y - (b + 2) = 0, provided that x ≠ 0. However, we should check if x = 0 is a root. Plugging x = 0 into the original equation: 0 + 0 - 0 + 0 + 1 = 1 ≠ 0, so x = 0 is not a root. Therefore, we can safely divide by x² without losing any roots.

Therefore, the original quartic equation has real roots if and only if the quadratic equation in y has real solutions y such that there exists some real x ≠ 0 with x + 1/x = y.

So, first, we need the quadratic equation in y to have real roots. Then, for each such real root y, we need to check if there exists a real x ≠ 0 such that x + 1/x = y.

So, let me first find conditions on a and b such that the quadratic equation y² + a y - (b + 2) = 0 has real solutions.

The discriminant of the quadratic equation in y is D = a² + 4(b + 2) = a² + 4b + 8. Since a and b are real numbers, this discriminant is always positive because a² ≥ 0 and 4b + 8 ≥ 4*0 + 8 = 8 > 0. Therefore, the quadratic equation in y always has two real roots for any a and b.

But even though the quadratic in y has real roots, we need to make sure that there exists a real x ≠ 0 such that x + 1/x = y.

So, let's analyze the equation x + 1/x = y. Let me solve for x:

Multiply both sides by x: x² - y x + 1 = 0. This quadratic equation in x has real solutions if and only if the discriminant is non-negative. The discriminant is y² - 4. Therefore, for real solutions x, we need y² - 4 ≥ 0 ⇒ |y| ≥ 2.

Therefore, for the original quartic equation to have a real root, there must exist a real solution y to the quadratic in y (which we know there are always two real solutions) such that |y| ≥ 2.

Therefore, the problem reduces to: find 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 real root with |y| ≥ 2.

So, now I need to find the conditions on a and b such that at least one of the roots y of y² + a y - (b + 2) = 0 satisfies |y| ≥ 2.

Let me denote the roots of the quadratic equation in y as y₁ and y₂. Since the quadratic is y² + a y - (b + 2) = 0, by Vieta's formula, we have:

y₁ + y₂ = -a

and

y₁ y₂ = - (b + 2)

Our goal is to find when at least one of y₁ or y₂ has absolute value ≥ 2.

Alternatively, we can think of this as the quadratic equation has at least one root with |y| ≥ 2.

To find when this happens, perhaps we can use the concept of quadratic equations and their roots.

Alternatively, we can think of it as follows:

For the quadratic equation y² + a y - (b + 2) = 0, if either y₁ ≥ 2 or y₂ ≤ -2 or y₁ ≤ -2 or y₂ ≥ 2.

But since Vieta's formulas relate the roots, maybe there's a way to translate these conditions into inequalities on a and b.

Alternatively, maybe we can use the fact that if a quadratic equation has at least one root with |y| ≥ 2, then either:

1. The quadratic has a root ≥ 2, or

2. The quadratic has a root ≤ -2

Therefore, we can find the conditions for each case and combine them.

Alternatively, we can use the following approach: For a quadratic equation f(y) = y² + a y - (b + 2) = 0, the condition that at least one root is ≥ 2 or ≤ -2 can be checked using the following method:

For a quadratic equation f(y) = 0 to have a root ≥ 2, either:

- The quadratic has a root at y = 2 (i.e., f(2) = 0), or

- The quadratic has a root greater than 2. Since the quadratic opens upwards (coefficient of y² is 1 > 0), then if f(2) ≤ 0, there is a root ≥ 2. Similarly, for a root ≤ -2, if f(-2) ≤ 0, there is a root ≤ -2.

Wait, let me recall that for a quadratic equation with leading coefficient positive, the quadratic is positive outside the interval between its roots and negative between them. Therefore, if f(2) ≤ 0, then there is a root ≥ 2, because the quadratic is positive at y approaching infinity and negative at y = 2. Similarly, f(-2) ≤ 0 would imply that there is a root ≤ -2.

Therefore, the condition that the quadratic equation has at least one root with |y| ≥ 2 is equivalent to f(2) ≤ 0 or f(-2) ≤ 0.

But wait, let me check that. Suppose that the quadratic has two roots, both between -2 and 2. Then f(2) and f(-2) would be positive? Let me think.

Wait, suppose the quadratic opens upwards and both roots are between -2 and 2. Then f(-2) > 0 and f(2) > 0. Conversely, if f(2) ≤ 0, then since the quadratic opens upwards, there must be a root ≥ 2. Similarly, if f(-2) ≤ 0, then there must be a root ≤ -2. Therefore, the condition that the quadratic has at least one root with |y| ≥ 2 is equivalent to f(2) ≤ 0 or f(-2) ≤ 0.

Therefore, our problem reduces to finding the set of (a, b) where either f(2) ≤ 0 or f(-2) ≤ 0.

Therefore, let me compute f(2) and f(-2):

First, 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.

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

Let me now write these inequalities:

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

Second inequality: 2 - 2a - b ≤ 0 ⇒ b ≥ 2 - 2a

But remember that a and b are in [0,1]. So let's analyze these two inequalities in the square [0,1]x[0,1].

First, for the first inequality: b ≥ 2 + 2a. Let's check if this is possible. Since a ∈ [0,1], 2 + 2a ranges from 2 + 0 = 2 to 2 + 2*1 = 4. But since b is at most 1, which is less than 2, the inequality b ≥ 2 + 2a can never be satisfied. Therefore, the first inequality is never true for (a, b) in [0,1]x[0,1]. Therefore, the only contributing inequality is the second one: b ≥ 2 - 2a.

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

Wait, but let me check again. The original condition was that the quadratic equation in y has at least one root with |y| ≥ 2, which is equivalent to f(2) ≤ 0 or f(-2) ≤ 0. We saw that f(2) ≤ 0 gives b ≥ 2 + 2a, which is impossible. Therefore, only f(-2) ≤ 0 gives b ≥ 2 - 2a. But let me check again if there's another scenario.

Wait, perhaps there's a mistake here. Because even if both roots are such that one is ≥ 2 or ≤ -2, but maybe the quadratic could have one root on one side and another on the other side? But according to the previous logic, if f(2) ≤ 0 or f(-2) ≤ 0, then there is a root on that side.

But perhaps there is another case where even if f(2) > 0 and f(-2) > 0, but one of the roots is still outside [-2, 2]? Wait, no. Because if the quadratic opens upwards and f(2) > 0 and f(-2) > 0, then both roots lie between -2 and 2. So, in that case, there are no roots with |y| ≥ 2. Therefore, the condition that there is at least one root outside [-2, 2] is equivalent to f(2) ≤ 0 or f(-2) ≤ 0.

Therefore, since the first inequality b ≥ 2 + 2a is impossible, only the second inequality b ≥ 2 - 2a is relevant.

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

But wait, let me confirm this with an example. Let me take a specific value of a and check. Let me take a = 0. Then the inequality becomes b ≥ 2 - 0 = 2. But since b is at most 1, there are no solutions here. However, when a = 0, let's check the original equation.

Original equation when a = 0: x⁴ - b x² + 1 = 0. Let me see if this equation has real roots. Let me set z = x², so z² - b z + 1 = 0. The discriminant is b² - 4. Therefore, real roots in z only if b² - 4 ≥ 0 ⇒ b ≥ 2 or b ≤ -2. Since b ∈ [0,1], there are no real roots for a = 0. Which matches with our conclusion that for a = 0, the inequality b ≥ 2 - 2a becomes b ≥ 2, which is impossible.

But then, when a = 1? Then the inequality becomes b ≥ 2 - 2(1) = 0. Therefore, for a = 1, all b ∈ [0,1] satisfy b ≥ 0. Therefore, all points (1, b) with b ∈ [0,1] are in S. Let me check if for a = 1 and any b, the original equation has a real root.

Original equation when a = 1: x⁴ + x³ - b x² + x + 1 = 0. Let me check if this has real roots.

Alternatively, let me use our previous analysis. For a = 1, the condition is b ≥ 0 (which is true since b ∈ [0,1]). Then, according to our previous logic, there should be a real root. Let me check for some specific b.

Take b = 0. Then the equation becomes x⁴ + x³ + x + 1 = 0. Let me factor this. Maybe factor by grouping: x³(x + 1) + 1(x + 1) = (x³ + 1)(x + 1). Wait, x³ + 1 = (x + 1)(x² - x + 1). Therefore, the equation is (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, there is a real root x = -1. Therefore, for a = 1, b = 0, there is a real root.

Another case: a = 1, b = 1. The equation is x⁴ + x³ - x² + x + 1 = 0. Let me check if this has real roots. Again, using substitution.

Let me try to see if x = -1 is a root: (-1)^4 + (-1)^3 - (-1)^2 + (-1) + 1 = 1 -1 -1 -1 +1 = -1 ≠ 0. How about x = -2? Let me check. (-2)^4 + (-2)^3 - (-2)^2 + (-2) + 1 = 16 - 8 - 4 - 2 + 1 = 3 ≠ 0. Maybe x = -1/2? Let me compute: (1/16) + (-1/8) - (1/4) + (-1/2) + 1. Let me compute:

1/16 - 2/16 - 4/16 - 8/16 + 16/16 = (1 - 2 -4 -8 +16)/16 = (3)/16 > 0.

Alternatively, maybe I can check the behavior of the function f(x) = x⁴ + x³ - x² + x + 1. For large positive x, f(x) tends to +infty. For large negative x, x⁴ dominates, so also tends to +infty. Let me check the value at x = 0: f(0) = 0 + 0 - 0 + 0 +1 = 1. At x = -1, f(-1) = 1 -1 -1 -1 +1 = -1. Therefore, since f(-1) = -1 and f(0) = 1, by Intermediate Value Theorem, there is a root between -1 and 0. Therefore, yes, there is a real root. Therefore, for a = 1, b = 1, there is a real root.

Therefore, this seems to confirm that for a = 1, all b are in S.

Therefore, our conclusion that the condition is b ≥ 2 - 2a seems correct. However, there's another thing to check: when 2 - 2a is less than or equal to 1, since b can go up to 1. Let me find for which a, 2 - 2a is less than or equal to 1.

Solve 2 - 2a ≤ 1 ⇒ -2a ≤ -1 ⇒ a ≥ 1/2.

Therefore, for a ≥ 1/2, 2 - 2a ≤ 1. For a < 1/2, 2 - 2a > 1. But since b can only go up to 1, for a < 1/2, 2 - 2a > 1, so the inequality b ≥ 2 - 2a would imply b ≥ something larger than 1, which is impossible. Therefore, for a < 1/2, there are no solutions.

Therefore, the set S is the set of (a, b) in [0,1]x[0,1] with a ≥ 1/2 and b ≥ 2 - 2a.

But let me check for a = 1/2. Then 2 - 2*(1/2) = 2 -1 = 1. So, the condition is b ≥ 1. Since b ∈ [0,1], that would mean b = 1. Therefore, the line b = 1 for a = 1/2.

Therefore, the region S is the set of points (a, b) where a is between 1/2 and 1, and b is between 2 - 2a and 1.

Therefore, to compute the area of S, we need to integrate over a from 1/2 to 1, the length of b from 2 - 2a to 1.

So the area is ∫ from a = 1/2 to 1 of [1 - (2 - 2a)] da = ∫ from 1/2 to 1 of (1 - 2 + 2a) da = ∫ from 1/2 to 1 of (2a -1) da

Let me compute that integral:

∫(2a -1) da = a² - a + C

Evaluated from 1/2 to 1:

At a = 1: 1² - 1 = 1 -1 = 0

At a = 1/2: ( (1/2)^2 - 1/2 ) = 1/4 - 1/2 = -1/4

Therefore, subtracting: 0 - (-1/4) = 1/4

Therefore, the area is 1/4? Wait, that seems small. Let me check.

Wait, the integral from a = 1/2 to 1 of (2a -1) da is equal to [a² - a] from 1/2 to 1.

At upper limit 1: 1 - 1 = 0

At lower limit 1/2: ( (1/2)^2 - 1/2 ) = 1/4 - 1/2 = -1/4.

Therefore, the integral is 0 - (-1/4) = 1/4. So yes, the area is 1/4.

But wait, let me think again. Because S is defined as points (a, b) where a ∈ [1/2, 1], b ∈ [2 - 2a, 1]. Therefore, the area should be the area under the line b = 2 - 2a from a = 1/2 to 1, subtracted from the area of the rectangle [1/2,1]x[0,1]. Wait, no. Wait, since for each a, b ranges from 2 - 2a up to 1. Therefore, the area is the integral over a from 1/2 to 1 of (upper bound - lower bound) da, which is ∫ (1 - (2 - 2a)) da from 1/2 to 1 = ∫(2a -1) da from 1/2 to 1, which we found to be 1/4. So 1/4 is the area?

But let me visualize this region. For a between 1/2 and 1, the lower bound for b is 2 - 2a. When a = 1/2, lower bound is 2 - 2*(1/2) = 1. So at a = 1/2, b is from 1 to 1, which is just the point (1/2, 1). As a increases from 1/2 to 1, the lower bound 2 - 2a decreases from 1 to 0. So the region is a triangle with vertices at (1/2, 1), (1, 1), and (1, 0). Wait, let me check.

Wait, when a = 1, 2 - 2a = 0. Therefore, the region is bounded by a = 1/2 to 1, b from 2 - 2a to 1. So plotting this in the a-b plane, it's a trapezoid? Or a triangle?

At a = 1/2, the lower bound is 1, upper bound is 1, so a vertical line at a=1/2, b=1. Then, as a increases, the lower bound descends from 1 to 0. At a=1, the lower bound is 0. So the region is a triangle with vertices at (1/2, 1), (1, 1), and (1, 0).

Because between a=1/2 and 1, the line b = 2 - 2a connects (1/2, 1) to (1, 0). So the area between this line, b=1, and a=1/2 to 1. So that is a triangle with base along b=1 from a=1/2 to 1, which is length 1 - 1/2 = 1/2. The height is from b=0 to b=1, but maybe better to calculate using coordinates.

Alternatively, since it's a triangle with vertices at (1/2, 1), (1, 1), and (1, 0). The area of this triangle can be calculated as (base * height)/2. The base is from (1/2, 1) to (1, 1), which is length 1 - 1/2 = 1/2. The height is from (1, 1) down to (1, 0), which is length 1. But wait, it's a right triangle? Wait, no. Wait, the three points are (1/2,1), (1,1), (1,0). So plotting these:

- The point (1/2,1) is the left point on the top edge.

- The point (1,1) is the top right corner.

- The point (1,0) is the bottom right corner.

So connecting these points, the figure is a right triangle with legs of length 1/2 (from a=1/2 to a=1) and 1 (from b=0 to b=1). Wait, no, actually, the base is along the top edge from (1/2,1) to (1,1), which is 1/2 units long. Then the vertical line from (1,1) down to (1,0) is 1 unit. However, the triangle formed is a right triangle with legs of length 1/2 and 1? Wait, actually, the triangle has vertices at (1/2,1), (1,1), and (1,0). So the base is between (1/2,1) and (1,1), which is length 1/2. The height is from (1,1) down to (1,0), which is length 1. Therefore, the area is (base * height)/2 = (1/2 * 1)/2 = 1/4. Which matches the integral result.

Therefore, the area of S is indeed 1/4. However, wait, but 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, S is only the triangle with area 1/4. But is that correct?

Wait, earlier when a = 0, there are no real roots. For a between 0 and 1/2, according to our analysis, there are no points (a, b) in S. However, perhaps this is incorrect? Maybe there are some cases where even though b ≥ 2 - 2a is not satisfied, there is still a real root.

Wait, let me think again. Our conclusion was that the condition is equivalent to b ≥ 2 - 2a. But how did we get there? Because we said that the quadratic in y has at least one root with |y| ≥ 2 iff f(2) ≤ 0 or f(-2) ≤ 0. We saw that f(2) ≤ 0 is impossible, so only f(-2) ≤ 0, which is b ≥ 2 - 2a. However, is there a possibility that even when f(-2) > 0 and f(2) > 0, there is a root with |y| ≥ 2?

Wait, earlier we thought that if f(-2) > 0 and f(2) > 0, then both roots are between -2 and 2. But is that necessarily the case?

Let me recall that for a quadratic equation with leading coefficient positive, if f(-2) > 0 and f(2) > 0, then both roots lie between -2 and 2?

Wait, not necessarily. Let me think. For example, suppose the quadratic has one root less than -2 and one root greater than 2. Then f(-2) would be negative (since the quadratic is positive at x approaching -infty and negative between the roots), and f(2) would be negative. Wait, no. Wait, if the quadratic opens upwards, and there is a root less than -2 and a root greater than 2, then f(-2) would be positive (since between -infty and the left root, the quadratic is positive) and f(2) would also be positive (since between the left root and the right root, the quadratic is negative, and then after the right root, it's positive again. Wait, if the left root is less than -2 and the right root is greater than 2, then at x = -2, which is between the left root and the right root, the quadratic would be negative. Therefore f(-2) < 0. Similarly, f(2) would be negative. Therefore, if there's a root outside of [-2, 2], then f(-2) or f(2) would be negative. Conversely, if both f(-2) and f(2) are positive, then both roots lie between -2 and 2.

So that logic holds. Therefore, our initial conclusion is correct: the original quartic has a real root iff b ≥ 2 - 2a. Therefore, S is the triangle with area 1/4.

But wait, earlier when we tested a = 1, which is in the region, and found that there are real roots, and when a = 1/2, b =1, there is a root. However, let me check for some a between 0 and 1/2, say a = 0. Let me check with a = 0.25, which is less than 1/2. Suppose b = 0. Let me check the original equation.

Original equation: x⁴ + (0.25)x³ - 0 x² + 0.25x + 1 = x⁴ + 0.25x³ + 0.25x + 1 = 0. Let me check if there are any real roots.

Again, trying to use substitution. Let me try dividing by x² (since x ≠ 0):

x² + 0.25x - 0 + 0.25/x + 1/x² = 0

Let me set y = x + 1/x again. Then x² + 1/x² = y² - 2. Therefore:

y² - 2 + 0.25 y = 0 ⇒ y² + 0.25 y - 2 = 0

Solving: y = [-0.25 ± sqrt( (0.25)^2 + 8 )]/2

Discriminant is 0.0625 + 8 = 8.0625 = (2.84)^2 approximately.

So the roots are y = [ -0.25 ± 2.84 ] / 2. Let me calculate approximate values:

First root: (-0.25 + 2.84)/2 ≈ 2.59 / 2 ≈ 1.295

Second root: (-0.25 - 2.84)/2 ≈ -3.09 / 2 ≈ -1.545

Now, we need to check if there exists real x with y = x + 1/x equal to 1.295 or -1.545.

For y = 1.295: the equation x + 1/x = 1.295. Multiply by x: x² - 1.295x + 1 = 0. Discriminant: (1.295)^2 - 4 ≈ 1.677 - 4 = -2.323 < 0. No real roots.

For y = -1.545: equation x + 1/x = -1.545. Multiply by x: x² + 1.545x + 1 = 0. Discriminant: (1.545)^2 - 4 ≈ 2.387 - 4 ≈ -1.613 < 0. Also no real roots. Therefore, there are no real roots for a = 0.25, b = 0.

Therefore, this seems to confirm that for a < 1/2, even with b = 0, there are no real roots. Hence, the region S is indeed only the triangle with area 1/4.

But let me check another point in the region. Suppose a = 3/4, which is greater than 1/2. Let's take b = 1. Then according to our condition, b = 1 ≥ 2 - 2*(3/4) = 2 - 3/2 = 1/2. So yes, it satisfies. Let me check if the equation has real roots.

Original equation: x⁴ + (3/4)x³ - 1 x² + (3/4)x + 1 = 0.

Again, use substitution. Divide by x²:

x² + (3/4)x - 1 + (3/4)/x + 1/x² = 0

Let y = x + 1/x. Then x² + 1/x² = y² - 2. Therefore:

y² - 2 + (3/4)y - 1 = 0 ⇒ y² + (3/4)y - 3 = 0

Solve for y: y = [-3/4 ± sqrt( (9/16) + 12 )]/2 = [-3/4 ± sqrt( 195/16 )]/2

sqrt(195)/4 ≈ 13.96/4 ≈ 3.49

Therefore, y ≈ [ -0.75 ± 3.49 ] / 2

First root: ( -0.75 + 3.49 ) / 2 ≈ 2.74 / 2 ≈ 1.37

Second root: ( -0.75 - 3.49 ) / 2 ≈ -4.24 / 2 ≈ -2.12

Check if these y values correspond to real x.

For y ≈ 1.37: x + 1/x = 1.37. Multiply by x: x² - 1.37x + 1 = 0. Discriminant: (1.37)^2 - 4 ≈ 1.88 - 4 = -2.12 < 0. No real roots.

For y ≈ -2.12: x + 1/x = -2.12. Multiply by x: x² + 2.12x + 1 = 0. Discriminant: (2.12)^2 - 4 ≈ 4.49 - 4 = 0.49 > 0. Therefore, real roots. Therefore, there is a real root in this case. Therefore, yes, the equation has real roots.

Therefore, this point (a=3/4, b=1) is in S, and the area calculation seems correct.

But let me check another point not in S. For example, a = 3/4, b = 0.5. Since 0.5 < 2 - 2*(3/4) = 2 - 1.5 = 0.5. Wait, 0.5 is equal to 2 - 2a, so this is on the boundary. Therefore, it should be in S. Let me check.

For a = 3/4, b = 0.5: original equation is x⁴ + (3/4)x³ - 0.5 x² + (3/4)x + 1 = 0. Let me check. Using substitution:

Divide by x²: x² + (3/4)x - 0.5 + (3/4)/x + 1/x² = 0

Let y = x + 1/x. Then x² + 1/x² = y² - 2. Therefore, equation becomes:

y² - 2 + (3/4)y - 0.5 = 0 ⇒ y² + (3/4)y - 2.5 = 0

Solutions: y = [-3/4 ± sqrt( (9/16) + 10 )]/2

Compute discriminant: 9/16 + 10 = 9/16 + 160/16 = 169/16. sqrt(169/16) = 13/4.

Therefore, y = [ -3/4 ± 13/4 ] / 2

First root: (-3/4 + 13/4)/2 = (10/4)/2 = (5/2)/2 = 5/4 = 1.25

Second root: (-3/4 -13/4)/2 = (-16/4)/2 = (-4)/2 = -2

So y = 1.25 and y = -2.

Check for y = -2: equation x + 1/x = -2. Multiply by x: x² + 2x + 1 = 0 ⇒ (x +1)^2 = 0 ⇒ x = -1. Therefore, x = -1 is a root.

Indeed, plugging x = -1 into original equation: (-1)^4 + (3/4)(-1)^3 - 0.5 (-1)^2 + (3/4)(-1) + 1 = 1 - 3/4 - 0.5 - 3/4 + 1 = 1 - 0.75 - 0.5 - 0.75 + 1 = (1 + 1) - (0.75 + 0.5 + 0.75) = 2 - 2 = 0. Therefore, x = -1 is a root. Therefore, this point is in S.

Therefore, our analysis holds.

Therefore, the area of S is 1/4.

Wait, but the problem says "the set of points (a,b) with 0 ≤ a, b ≤ 1 such that the equation has at least one real root". However, according to our analysis, the area is 1/4, but I want to make sure there isn't something missing.

Wait, could there be cases where even though b < 2 - 2a, the original equation still has a real root?

Let me take a point not in S, say a = 3/4, b = 0.4. Since 2 - 2*(3/4) = 0.5, and 0.4 < 0.5, so this point is not in S. Let me check if the equation has real roots.

Original equation: x⁴ + (3/4)x³ - 0.4x² + (3/4)x + 1 = 0.

Again, using substitution. Divide by x²:

x² + (3/4)x - 0.4 + (3/4)/x + 1/x² = 0

Let y = x + 1/x. Then x² + 1/x² = y² - 2.

Equation becomes: y² - 2 + (3/4)y - 0.4 = 0 ⇒ y² + (3/4)y - 2.4 = 0

Solve for y: y = [ -3/4 ± sqrt( (3/4)^2 + 9.6 ) ] / 2

Compute discriminant: (9/16) + 9.6 = 9/16 + 153.6/16 = 162.6/16 ≈ 10.1625

sqrt(10.1625) ≈ 3.19

Therefore, y ≈ [ -0.75 ± 3.19 ] / 2

First root: (-0.75 + 3.19)/2 ≈ 2.44 / 2 ≈ 1.22

Second root: (-0.75 - 3.19)/2 ≈ -3.94 / 2 ≈ -1.97

Check for real x:

For y ≈ 1.22: equation x + 1/x = 1.22. Multiply by x: x² - 1.22x + 1 = 0. Discriminant ≈ 1.4884 - 4 ≈ -2.5116 < 0. No real roots.

For y ≈ -1.97: equation x + 1/x = -1.97. Multiply by x: x² + 1.97x + 1 = 0. Discriminant ≈ (1.97)^2 - 4 ≈ 3.88 - 4 ≈ -0.12 < 0. Therefore, no real roots. Hence, no real roots for this point.

Therefore, the area calculation seems correct.

Therefore, the answer is 1/4.

But wait, I remember that sometimes when you do these substitutions, you might miss some roots? Let me think again.

We started with the quartic equation, divided by x², substituted y = x + 1/x, and transformed into a quadratic in y. Then, we determined that for the quartic to have a real root, the quadratic must have a root with |y| ≥ 2.

But suppose that x = 1 or x = -1? Let me check if these can be roots.

For example, if x = 1: Plug into the original equation: 1 + a - b + a + 1 = 2 + 2a - b. So 2 + 2a - b = 0. Therefore, if 2 + 2a - b = 0, then x = 1 is a root. Similarly, x = -1: (-1)^4 + a(-1)^3 - b(-1)^2 + a(-1) + 1 = 1 -a - b -a + 1 = 2 - 2a - b. Therefore, if 2 - 2a - b = 0, then x = -1 is a root.

Wait, so these are specific cases. But in our previous analysis, we considered that the quartic has a real root when either f(2) ≤ 0 or f(-2) ≤ 0. Which translates to 2 + 2a - b ≤ 0 or 2 - 2a - b ≤ 0.

But if 2 + 2a - b = 0, then x = 1 is a root, so our condition includes that case. Similarly, 2 - 2a - b = 0 implies x = -1 is a root.

Therefore, our analysis includes those cases.

But in our region S, we only considered the condition b ≥ 2 - 2a. But when b = 2 - 2a, then 2 - 2a - b = 0, which is exactly when x = -1 is a root. Therefore, the boundary of our region S is exactly when x = -1 is a root.

Therefore, our region S includes exactly those points (a, b) where the original equation has a real root. Therefore, the area is indeed 1/4.

But let me check if there are other real roots which are not captured by this analysis. For example, suppose there's a real root x ≠ ±1. But according to our substitution, any real root x ≠ 0 would lead to a solution y with |y| ≥ 2. Therefore, our analysis should capture all possibilities.

Therefore, the answer should be 1/4.

But let me check once more with another example. Let me take a = 1/2, b = 1. As we saw earlier, the equation has a root at x = -1. Let me check. Plug x = -1 into the equation: (-1)^4 + (1/2)(-1)^3 - 1*(-1)^2 + (1/2)(-1) + 1 = 1 - 1/2 -1 -1/2 +1 = 1 - 0.5 -1 -0.5 +1 = (1 + 1) - (0.5 + 1 + 0.5) = 2 - 2 = 0. Yes, it works.

Another case: a = 1, b = 1/2. Then according to the region, since a = 1 ≥ 1/2, and b = 1/2 ≥ 2 - 2*1 = 0. So yes. Let me check if the equation has a real root.

Original equation: x⁴ + x³ - (1/2)x² + x + 1 = 0.

Again, using substitution. Divide by x²:

x² + x - 1/2 + 1/x + 1/x² = 0

Let y = x + 1/x, so x² + 1/x² = y² - 2. Then equation becomes:

y² - 2 + x - 1/2 + 1/x = 0. Wait, no, original equation:

Wait, the original equation after dividing by x² is x² + x - 1/2 + 1/x + 1/x² = 0. Which is (x² + 1/x²) + x + 1/x - 1/2 = 0.

So that is (y² - 2) + y - 1/2 = 0 ⇒ y² + y - 2.5 = 0.

Solving: y = [-1 ± sqrt(1 + 10)] / 2 = [-1 ± sqrt(11)] / 2 ≈ [-1 ± 3.3166]/2.

First root: (-1 + 3.3166)/2 ≈ 2.3166 / 2 ≈ 1.1583

Second root: (-1 - 3.3166)/2 ≈ -4.3166 / 2 ≈ -2.1583

Check for the second root, y ≈ -2.1583. Then equation x + 1/x = -2.1583. Multiply by x: x² + 2.1583x + 1 = 0. Discriminant ≈ (2.1583)^2 - 4 ≈ 4.658 - 4 = 0.658 > 0. Therefore, real roots. Therefore, there is a real root.

Hence, the point (a=1, b=1/2) is in S, which is correct.

Therefore, after thorough checking, I believe that the area of S is indeed 1/4.

But just to be thorough, let me think if there's another approach. For example, suppose we consider the original quartic equation.

We can analyze the function f(x) = x⁴ + a x³ - b x² + a x + 1. We want to know for which (a,b) there exists x ∈ ℝ such that f(x) = 0.

Since f(x) is a continuous function, we can consider its minimum value. If the minimum value of f(x) is ≤ 0, then there's a real root.

But finding the minimum of a quartic is complicated. However, given the symmetry we found earlier, maybe we can use it.

Alternatively, considering the substitution we did earlier, which reduced the problem to analyzing y.

But our previous approach seems valid, and all test cases confirm that.

Therefore, I think confidently that the area is 1/4.

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

To solve the problem, we are given a quartic polynomial:

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

and asked to determine the **area** of the region $ S $ in the unit square $ 0 \leq a, b \leq 1 $ such that this polynomial has **at least one real root**. The goal is to find the set $ S $ and then compute its area.

---

### Step 1: Analyze the Equation

We make the substitution $ y = x + \frac{1}{x} $. This is useful because the polynomial is **symmetric** around the middle term, and dividing through by $ x^2 $ (valid when $ x \neq 0 $) gives:

$$
x^4 + ax^3 - bx^2 + ax + 1 = 0 \Rightarrow x^2 + a x - b + \frac{a}{x} + \frac{1}{x^2} = 0
$$

Rewriting in terms of $ y = x + \frac{1}{x} $, we can use:
$$
x^2 + \frac{1}{x^2} = y^2 - 2
$$
So the equation becomes:
$$
y^2 - 2 + a y - b = 0 \Rightarrow y^2 + a y - (b + 2) = 0
$$

This is a **quadratic in $ y $**. For the original equation to have any **real roots** in $ x $, some real solution $ y $ must be such that $ x + \frac{1}{x} = y $ has a **real solution** $ x $.

This is guaranteed to have real solutions $ x \neq 0 $ **if and only if**:

$$
|y| \geq 2
$$

---

### Step 2: Characterizing the Set $ S $

We now find the set of $ (a, b) $ such that the quadratic equation $ y^2 + a y - (b + 2) = 0 $ has **at least one real root** with $ |y| \geq 2 $.

Let’s denote the two roots of the quadratic as $ y_1 $ and $ y_2 $. Since the quadratic has real roots, it is always true (as $ a, b \in [0,1] \Rightarrow a^2 + 4(b+2) > 0 $), and we only need to check if **at least one root** satisfies $ |y| \geq 2 $.

We use the **test values** at $ y = 2 $ and $ y = -2 $, since the sign of the quadratic function at these points can characterize the root locations.

Let:

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

Then:

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

We consider cases where there is at least one real root with $ |y| \geq 2 $, which is guaranteed if:

$$
f(2) \leq 0 \quad \text{or} \quad f(-2) \leq 0
$$

We are to find the region in $ 0 \leq a, b \leq 1 $ such that at least one of these is true.

---

### Step 3: Determine the Region $ S $

We analyze:

1. $ f(2) \leq 0 \Rightarrow 2 + 2a - b \leq 0 \Rightarrow b \geq 2 + 2a $

   Since $ a \in [0, 1] $, $ 2 + 2a \in [2, 4] $, and $ b \in [0, 1] $, this condition **cannot ever** hold.

2. $ f(-2) \leq 0 \Rightarrow 2 - 2a - b \leq 0 \Rightarrow b \geq 2 - 2a $

This condition defines a **region** in the unit square $ [0,1] \times [0,1] $, and since $ a \in [0,1] $, we find:

- $ 2 - 2a \in [0, 2] $, so $ b \geq 2 - 2a $ is possible only when $ b \in [2 - 2a, 1] $, as $ b \leq 1 $

Thus, the set $ S $ is defined by:

$$
S = \left\{ (a, b) \in [0,1] \times [0,1] \mid b \geq 2 - 2a \right\}
$$

This is a **triangle** in the $ a $-$ b $ plane, bounded by:

- $ b = 2 - 2a $, the line from $ (a, b) = (1, 0) $ to $ (1/2, 1) $
- $ a = 1/2 $, where $ b = 1 $
- $ a = 1 $, where $ b = 0 $
- and the vertical lines $ a = 0 $ and $ b = 0 $, but only the triangle defined above is included in the region.

This triangle has:

- Base from $ a = 1/2 $ to $ a = 1 $, length $ 1/2 $
- Height from $ b = 0 $ to $ b = 1 $, length $ 1 $

So the area is:

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

---

### Final Answer

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

[Assistant]
Explanation:
Okay, so I need to find the area of the set S, which consists of all 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. Alright, let me start by understanding the problem.

First, the equation given is a quartic polynomial: x⁴ + a x³ - b x² + a x + 1 = 0. The coefficients a and b are between 0 and 1. The question is asking for the area of all (a, b) pairs where this equation has at least one real root.

So, my goal is to figure out the region in the [0,1]x[0,1] square where this polynomial has at least one real root, and then compute the area of that region.

Let me recall that for polynomials, especially quartic ones, it's not straightforward to determine the number of real roots. However, maybe there's a way to simplify this equation or find some conditions on a and b that would guarantee a real root.

Let me first check if there are any obvious substitutions or symmetries. The polynomial is x⁴ + a x³ - b x² + a x + 1. If I look at the coefficients, the coefficients of x⁴ and the constant term are both 1. Also, the coefficients of x³ and x are both a. That symmetry might be useful.

Perhaps I can factor this polynomial or rewrite it in a way that makes it easier to analyze. Let me try to see if it can be written as a quadratic in terms of x + 1/x or something like that.

Let me consider dividing both sides by x² to make it symmetric. Let me try that:

Dividing by x²:

x² + a x - b + a / x + 1 / x² = 0

Hmm, let me rewrite this:

x² + 1/x² + a(x + 1/x) - b = 0

Yes, that seems promising. Let me set y = x + 1/x. Then, note that y² = x² + 2 + 1/x², so x² + 1/x² = y² - 2.

Therefore, substituting into the equation:

(y² - 2) + a y - b = 0 ⇒ y² + a y - (b + 2) = 0

So, the original quartic equation is equivalent to the quadratic equation in y: y² + a y - (b + 2) = 0, provided that x ≠ 0. However, we should check if x = 0 is a root. Plugging x = 0 into the original equation: 0 + 0 - 0 + 0 + 1 = 1 ≠ 0, so x = 0 is not a root. Therefore, we can safely divide by x² without losing any roots.

Therefore, the original quartic equation has real roots if and only if the quadratic equation in y has real solutions y such that there exists some real x ≠ 0 with x + 1/x = y.

So, first, we need the quadratic equation in y to have real roots. Then, for each such real root y, we need to check if there exists a real x ≠ 0 such that x + 1/x = y.

So, let me first find conditions on a and b such that the quadratic equation y² + a y - (b + 2) = 0 has real solutions.

The discriminant of the quadratic equation in y is D = a² + 4(b + 2) = a² + 4b + 8. Since a and b are real numbers, this discriminant is always positive because a² ≥ 0 and 4b + 8 ≥ 4*0 + 8 = 8 > 0. Therefore, the quadratic equation in y always has two real roots for any a and b.

But even though the quadratic in y has real roots, we need to make sure that there exists a real x ≠ 0 such that x + 1/x = y.

So, let's analyze the equation x + 1/x = y. Let me solve for x:

Multiply both sides by x: x² - y x + 1 = 0. This quadratic equation in x has real solutions if and only if the discriminant is non-negative. The discriminant is y² - 4. Therefore, for real solutions x, we need y² - 4 ≥ 0 ⇒ |y| ≥ 2.

Therefore, for the original quartic equation to have a real root, there must exist a real solution y to the quadratic in y (which we know there are always two real solutions) such that |y| ≥ 2.

Therefore, the problem reduces to: find 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 real root with |y| ≥ 2.

So, now I need to find the conditions on a and b such that at least one of the roots y of y² + a y - (b + 2) = 0 satisfies |y| ≥ 2.

Let me denote the roots of the quadratic equation in y as y₁ and y₂. Since the quadratic is y² + a y - (b + 2) = 0, by Vieta's formula, we have:

y₁ + y₂ = -a

and

y₁ y₂ = - (b + 2)

Our goal is to find when at least one of y₁ or y₂ has absolute value ≥ 2.

Alternatively, we can think of this as the quadratic equation has at least one root with |y| ≥ 2.

To find when this happens, perhaps we can use the concept of quadratic equations and their roots.

Alternatively, we can think of it as follows:

For the quadratic equation y² + a y - (b + 2) = 0, if either y₁ ≥ 2 or y₂ ≤ -2 or y₁ ≤ -2 or y₂ ≥ 2.

But since Vieta's formulas relate the roots, maybe there's a way to translate these conditions into inequalities on a and b.

Alternatively, maybe we can use the fact that if a quadratic equation has at least one root with |y| ≥ 2, then either:

1. The quadratic has a root ≥ 2, or

2. The quadratic has a root ≤ -2

Therefore, we can find the conditions for each case and combine them.

Alternatively, we can use the following approach: For a quadratic equation f(y) = y² + a y - (b + 2) = 0, the condition that at least one root is ≥ 2 or ≤ -2 can be checked using the following method:

For a quadratic equation f(y) = 0 to have a root ≥ 2, either:

- The quadratic has a root at y = 2 (i.e., f(2) = 0), or

- The quadratic has a root greater than 2. Since the quadratic opens upwards (coefficient of y² is 1 > 0), then if f(2) ≤ 0, there is a root ≥ 2. Similarly, for a root ≤ -2, if f(-2) ≤ 0, there is a root ≤ -2.

Wait, let me recall that for a quadratic equation with leading coefficient positive, the quadratic is positive outside the interval between its roots and negative between them. Therefore, if f(2) ≤ 0, then there is a root ≥ 2, because the quadratic is positive at y approaching infinity and negative at y = 2. Similarly, f(-2) ≤ 0 would imply that there is a root ≤ -2.

Therefore, the condition that the quadratic equation has at least one root with |y| ≥ 2 is equivalent to f(2) ≤ 0 or f(-2) ≤ 0.

But wait, let me check that. Suppose that the quadratic has two roots, both between -2 and 2. Then f(2) and f(-2) would be positive? Let me think.

Wait, suppose the quadratic opens upwards and both roots are between -2 and 2. Then f(-2) > 0 and f(2) > 0. Conversely, if f(2) ≤ 0, then since the quadratic opens upwards, there must be a root ≥ 2. Similarly, if f(-2) ≤ 0, then there must be a root ≤ -2. Therefore, the condition that the quadratic has at least one root with |y| ≥ 2 is equivalent to f(2) ≤ 0 or f(-2) ≤ 0.

Therefore, our problem reduces to finding the set of (a, b) where either f(2) ≤ 0 or f(-2) ≤ 0.

Therefore, let me compute f(2) and f(-2):

First, 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.

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

Let me now write these inequalities:

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

Second inequality: 2 - 2a - b ≤ 0 ⇒ b ≥ 2 - 2a

But remember that a and b are in [0,1]. So let's analyze these two inequalities in the square [0,1]x[0,1].

First, for the first inequality: b ≥ 2 + 2a. Let's check if this is possible. Since a ∈ [0,1], 2 + 2a ranges from 2 + 0 = 2 to 2 + 2*1 = 4. But since b is at most 1, which is less than 2, the inequality b ≥ 2 + 2a can never be satisfied. Therefore, the first inequality is never true for (a, b) in [0,1]x[0,1]. Therefore, the only contributing inequality is the second one: b ≥ 2 - 2a.

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

Wait, but let me check again. The original condition was that the quadratic equation in y has at least one root with |y| ≥ 2, which is equivalent to f(2) ≤ 0 or f(-2) ≤ 0. We saw that f(2) ≤ 0 gives b ≥ 2 + 2a, which is impossible. Therefore, only f(-2) ≤ 0 gives b ≥ 2 - 2a. But let me check again if there's another scenario.

Wait, perhaps there's a mistake here. Because even if both roots are such that one is ≥ 2 or ≤ -2, but maybe the quadratic could have one root on one side and another on the other side? But according to the previous logic, if f(2) ≤ 0 or f(-2) ≤ 0, then there is a root on that side.

But perhaps there is another case where even if f(2) > 0 and f(-2) > 0, but one of the roots is still outside [-2, 2]? Wait, no. Because if the quadratic opens upwards and f(2) > 0 and f(-2) > 0, then both roots lie between -2 and 2. So, in that case, there are no roots with |y| ≥ 2. Therefore, the condition that there is at least one root outside [-2, 2] is equivalent to f(2) ≤ 0 or f(-2) ≤ 0.

Therefore, since the first inequality b ≥ 2 + 2a is impossible, only the second inequality b ≥ 2 - 2a is relevant.

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

But wait, let me confirm this with an example. Let me take a specific value of a and check. Let me take a = 0. Then the inequality becomes b ≥ 2 - 0 = 2. But since b is at most 1, there are no solutions here. However, when a = 0, let's check the original equation.

Original equation when a = 0: x⁴ - b x² + 1 = 0. Let me see if this equation has real roots. Let me set z = x², so z² - b z + 1 = 0. The discriminant is b² - 4. Therefore, real roots in z only if b² - 4 ≥ 0 ⇒ b ≥ 2 or b ≤ -2. Since b ∈ [0,1], there are no real roots for a = 0. Which matches with our conclusion that for a = 0, the inequality b ≥ 2 - 2a becomes b ≥ 2, which is impossible.

But then, when a = 1? Then the inequality becomes b ≥ 2 - 2(1) = 0. Therefore, for a = 1, all b ∈ [0,1] satisfy b ≥ 0. Therefore, all points (1, b) with b ∈ [0,1] are in S. Let me check if for a = 1 and any b, the original equation has a real root.

Original equation when a = 1: x⁴ + x³ - b x² + x + 1 = 0. Let me check if this has real roots.

Alternatively, let me use our previous analysis. For a = 1, the condition is b ≥ 0 (which is true since b ∈ [0,1]). Then, according to our previous logic, there should be a real root. Let me check for some specific b.

Take b = 0. Then the equation becomes x⁴ + x³ + x + 1 = 0. Let me factor this. Maybe factor by grouping: x³(x + 1) + 1(x + 1) = (x³ + 1)(x + 1). Wait, x³ + 1 = (x + 1)(x² - x + 1). Therefore, the equation is (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, there is a real root x = -1. Therefore, for a = 1, b = 0, there is a real root.

Another case: a = 1, b = 1. The equation is x⁴ + x³ - x² + x + 1 = 0. Let me check if this has real roots. Again, using substitution.

Let me try to see if x = -1 is a root: (-1)^4 + (-1)^3 - (-1)^2 + (-1) + 1 = 1 -1 -1 -1 +1 = -1 ≠ 0. How about x = -2? Let me check. (-2)^4 + (-2)^3 - (-2)^2 + (-2) + 1 = 16 - 8 - 4 - 2 + 1 = 3 ≠ 0. Maybe x = -1/2? Let me compute: (1/16) + (-1/8) - (1/4) + (-1/2) + 1. Let me compute:

1/16 - 2/16 - 4/16 - 8/16 + 16/16 = (1 - 2 -4 -8 +16)/16 = (3)/16 > 0.

Alternatively, maybe I can check the behavior of the function f(x) = x⁴ + x³ - x² + x + 1. For large positive x, f(x) tends to +infty. For large negative x, x⁴ dominates, so also tends to +infty. Let me check the value at x = 0: f(0) = 0 + 0 - 0 + 0 +1 = 1. At x = -1, f(-1) = 1 -1 -1 -1 +1 = -1. Therefore, since f(-1) = -1 and f(0) = 1, by Intermediate Value Theorem, there is a root between -1 and 0. Therefore, yes, there is a real root. Therefore, for a = 1, b = 1, there is a real root.

Therefore, this seems to confirm that for a = 1, all b are in S.

Therefore, our conclusion that the condition is b ≥ 2 - 2a seems correct. However, there's another thing to check: when 2 - 2a is less than or equal to 1, since b can go up to 1. Let me find for which a, 2 - 2a is less than or equal to 1.

Solve 2 - 2a ≤ 1 ⇒ -2a ≤ -1 ⇒ a ≥ 1/2.

Therefore, for a ≥ 1/2, 2 - 2a ≤ 1. For a < 1/2, 2 - 2a > 1. But since b can only go up to 1, for a < 1/2, 2 - 2a > 1, so the inequality b ≥ 2 - 2a would imply b ≥ something larger than 1, which is impossible. Therefore, for a < 1/2, there are no solutions.

Therefore, the set S is the set of (a, b) in [0,1]x[0,1] with a ≥ 1/2 and b ≥ 2 - 2a.

But let me check for a = 1/2. Then 2 - 2*(1/2) = 2 -1 = 1. So, the condition is b ≥ 1. Since b ∈ [0,1], that would mean b = 1. Therefore, the line b = 1 for a = 1/2.

Therefore, the region S is the set of points (a, b) where a is between 1/2 and 1, and b is between 2 - 2a and 1.

Therefore, to compute the area of S, we need to integrate over a from 1/2 to 1, the length of b from 2 - 2a to 1.

So the area is ∫ from a = 1/2 to 1 of [1 - (2 - 2a)] da = ∫ from 1/2 to 1 of (1 - 2 + 2a) da = ∫ from 1/2 to 1 of (2a -1) da

Let me compute that integral:

∫(2a -1) da = a² - a + C

Evaluated from 1/2 to 1:

At a = 1: 1² - 1 = 1 -1 = 0

At a = 1/2: ( (1/2)^2 - 1/2 ) = 1/4 - 1/2 = -1/4

Therefore, subtracting: 0 - (-1/4) = 1/4

Therefore, the area is 1/4? Wait, that seems small. Let me check.

Wait, the integral from a = 1/2 to 1 of (2a -1) da is equal to [a² - a] from 1/2 to 1.

At upper limit 1: 1 - 1 = 0

At lower limit 1/2: ( (1/2)^2 - 1/2 ) = 1/4 - 1/2 = -1/4.

Therefore, the integral is 0 - (-1/4) = 1/4. So yes, the area is 1/4.

But wait, let me think again. Because S is defined as points (a, b) where a ∈ [1/2, 1], b ∈ [2 - 2a, 1]. Therefore, the area should be the area under the line b = 2 - 2a from a = 1/2 to 1, subtracted from the area of the rectangle [1/2,1]x[0,1]. Wait, no. Wait, since for each a, b ranges from 2 - 2a up to 1. Therefore, the area is the integral over a from 1/2 to 1 of (upper bound - lower bound) da, which is ∫ (1 - (2 - 2a)) da from 1/2 to 1 = ∫(2a -1) da from 1/2 to 1, which we found to be 1/4. So 1/4 is the area?

But let me visualize this region. For a between 1/2 and 1, the lower bound for b is 2 - 2a. When a = 1/2, lower bound is 2 - 2*(1/2) = 1. So at a = 1/2, b is from 1 to 1, which is just the point (1/2, 1). As a increases from 1/2 to 1, the lower bound 2 - 2a decreases from 1 to 0. So the region is a triangle with vertices at (1/2, 1), (1, 1), and (1, 0). Wait, let me check.

Wait, when a = 1, 2 - 2a = 0. Therefore, the region is bounded by a = 1/2 to 1, b from 2 - 2a to 1. So plotting this in the a-b plane, it's a trapezoid? Or a triangle?

At a = 1/2, the lower bound is 1, upper bound is 1, so a vertical line at a=1/2, b=1. Then, as a increases, the lower bound descends from 1 to 0. At a=1, the lower bound is 0. So the region is a triangle with vertices at (1/2, 1), (1, 1), and (1, 0).

Because between a=1/2 and 1, the line b = 2 - 2a connects (1/2, 1) to (1, 0). So the area between this line, b=1, and a=1/2 to 1. So that is a triangle with base along b=1 from a=1/2 to 1, which is length 1 - 1/2 = 1/2. The height is from b=0 to b=1, but maybe better to calculate using coordinates.

Alternatively, since it's a triangle with vertices at (1/2, 1), (1, 1), and (1, 0). The area of this triangle can be calculated as (base * height)/2. The base is from (1/2, 1) to (1, 1), which is length 1 - 1/2 = 1/2. The height is from (1, 1) down to (1, 0), which is length 1. But wait, it's a right triangle? Wait, no. Wait, the three points are (1/2,1), (1,1), (1,0). So plotting these:

- The point (1/2,1) is the left point on the top edge.

- The point (1,1) is the top right corner.

- The point (1,0) is the bottom right corner.

So connecting these points, the figure is a right triangle with legs of length 1/2 (from a=1/2 to a=1) and 1 (from b=0 to b=1). Wait, no, actually, the base is along the top edge from (1/2,1) to (1,1), which is 1/2 units long. Then the vertical line from (1,1) down to (1,0) is 1 unit. However, the triangle formed is a right triangle with legs of length 1/2 and 1? Wait, actually, the triangle has vertices at (1/2,1), (1,1), and (1,0). So the base is between (1/2,1) and (1,1), which is length 1/2. The height is from (1,1) down to (1,0), which is length 1. Therefore, the area is (base * height)/2 = (1/2 * 1)/2 = 1/4. Which matches the integral result.

Therefore, the area of S is indeed 1/4. However, wait, but 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, S is only the triangle with area 1/4. But is that correct?

Wait, earlier when a = 0, there are no real roots. For a between 0 and 1/2, according to our analysis, there are no points (a, b) in S. However, perhaps this is incorrect? Maybe there are some cases where even though b ≥ 2 - 2a is not satisfied, there is still a real root.

Wait, let me think again. Our conclusion was that the condition is equivalent to b ≥ 2 - 2a. But how did we get there? Because we said that the quadratic in y has at least one root with |y| ≥ 2 iff f(2) ≤ 0 or f(-2) ≤ 0. We saw that f(2) ≤ 0 is impossible, so only f(-2) ≤ 0, which is b ≥ 2 - 2a. However, is there a possibility that even when f(-2) > 0 and f(2) > 0, there is a root with |y| ≥ 2?

Wait, earlier we thought that if f(-2) > 0 and f(2) > 0, then both roots are between -2 and 2. But is that necessarily the case?

Let me recall that for a quadratic equation with leading coefficient positive, if f(-2) > 0 and f(2) > 0, then both roots lie between -2 and 2?

Wait, not necessarily. Let me think. For example, suppose the quadratic has one root less than -2 and one root greater than 2. Then f(-2) would be negative (since the quadratic is positive at x approaching -infty and negative between the roots), and f(2) would be negative. Wait, no. Wait, if the quadratic opens upwards, and there is a root less than -2 and a root greater than 2, then f(-2) would be positive (since between -infty and the left root, the quadratic is positive) and f(2) would also be positive (since between the left root and the right root, the quadratic is negative, and then after the right root, it's positive again. Wait, if the left root is less than -2 and the right root is greater than 2, then at x = -2, which is between the left root and the right root, the quadratic would be negative. Therefore f(-2) < 0. Similarly, f(2) would be negative. Therefore, if there's a root outside of [-2, 2], then f(-2) or f(2) would be negative. Conversely, if both f(-2) and f(2) are positive, then both roots lie between -2 and 2.

So that logic holds. Therefore, our initial conclusion is correct: the original quartic has a real root iff b ≥ 2 - 2a. Therefore, S is the triangle with area 1/4.

But wait, earlier when we tested a = 1, which is in the region, and found that there are real roots, and when a = 1/2, b =1, there is a root. However, let me check for some a between 0 and 1/2, say a = 0. Let me check with a = 0.25, which is less than 1/2. Suppose b = 0. Let me check the original equation.

Original equation: x⁴ + (0.25)x³ - 0 x² + 0.25x + 1 = x⁴ + 0.25x³ + 0.25x + 1 = 0. Let me check if there are any real roots.

Again, trying to use substitution. Let me try dividing by x² (since x ≠ 0):

x² + 0.25x - 0 + 0.25/x + 1/x² = 0

Let me set y = x + 1/x again. Then x² + 1/x² = y² - 2. Therefore:

y² - 2 + 0.25 y = 0 ⇒ y² + 0.25 y - 2 = 0

Solving: y = [-0.25 ± sqrt( (0.25)^2 + 8 )]/2

Discriminant is 0.0625 + 8 = 8.0625 = (2.84)^2 approximately.

So the roots are y = [ -0.25 ± 2.84 ] / 2. Let me calculate approximate values:

First root: (-0.25 + 2.84)/2 ≈ 2.59 / 2 ≈ 1.295

Second root: (-0.25 - 2.84)/2 ≈ -3.09 / 2 ≈ -1.545

Now, we need to check if there exists real x with y = x + 1/x equal to 1.295 or -1.545.

For y = 1.295: the equation x + 1/x = 1.295. Multiply by x: x² - 1.295x + 1 = 0. Discriminant: (1.295)^2 - 4 ≈ 1.677 - 4 = -2.323 < 0. No real roots.

For y = -1.545: equation x + 1/x = -1.545. Multiply by x: x² + 1.545x + 1 = 0. Discriminant: (1.545)^2 - 4 ≈ 2.387 - 4 ≈ -1.613 < 0. Also no real roots. Therefore, there are no real roots for a = 0.25, b = 0.

Therefore, this seems to confirm that for a < 1/2, even with b = 0, there are no real roots. Hence, the region S is indeed only the triangle with area 1/4.

But let me check another point in the region. Suppose a = 3/4, which is greater than 1/2. Let's take b = 1. Then according to our condition, b = 1 ≥ 2 - 2*(3/4) = 2 - 3/2 = 1/2. So yes, it satisfies. Let me check if the equation has real roots.

Original equation: x⁴ + (3/4)x³ - 1 x² + (3/4)x + 1 = 0.

Again, use substitution. Divide by x²:

x² + (3/4)x - 1 + (3/4)/x + 1/x² = 0

Let y = x + 1/x. Then x² + 1/x² = y² - 2. Therefore:

y² - 2 + (3/4)y - 1 = 0 ⇒ y² + (3/4)y - 3 = 0

Solve for y: y = [-3/4 ± sqrt( (9/16) + 12 )]/2 = [-3/4 ± sqrt( 195/16 )]/2

sqrt(195)/4 ≈ 13.96/4 ≈ 3.49

Therefore, y ≈ [ -0.75 ± 3.49 ] / 2

First root: ( -0.75 + 3.49 ) / 2 ≈ 2.74 / 2 ≈ 1.37

Second root: ( -0.75 - 3.49 ) / 2 ≈ -4.24 / 2 ≈ -2.12

Check if these y values correspond to real x.

For y ≈ 1.37: x + 1/x = 1.37. Multiply by x: x² - 1.37x + 1 = 0. Discriminant: (1.37)^2 - 4 ≈ 1.88 - 4 = -2.12 < 0. No real roots.

For y ≈ -2.12: x + 1/x = -2.12. Multiply by x: x² + 2.12x + 1 = 0. Discriminant: (2.12)^2 - 4 ≈ 4.49 - 4 = 0.49 > 0. Therefore, real roots. Therefore, there is a real root in this case. Therefore, yes, the equation has real roots.

Therefore, this point (a=3/4, b=1) is in S, and the area calculation seems correct.

But let me check another point not in S. For example, a = 3/4, b = 0.5. Since 0.5 < 2 - 2*(3/4) = 2 - 1.5 = 0.5. Wait, 0.5 is equal to 2 - 2a, so this is on the boundary. Therefore, it should be in S. Let me check.

For a = 3/4, b = 0.5: original equation is x⁴ + (3/4)x³ - 0.5 x² + (3/4)x + 1 = 0. Let me check. Using substitution:

Divide by x²: x² + (3/4)x - 0.5 + (3/4)/x + 1/x² = 0

Let y = x + 1/x. Then x² + 1/x² = y² - 2. Therefore, equation becomes:

y² - 2 + (3/4)y - 0.5 = 0 ⇒ y² + (3/4)y - 2.5 = 0

Solutions: y = [-3/4 ± sqrt( (9/16) + 10 )]/2

Compute discriminant: 9/16 + 10 = 9/16 + 160/16 = 169/16. sqrt(169/16) = 13/4.

Therefore, y = [ -3/4 ± 13/4 ] / 2

First root: (-3/4 + 13/4)/2 = (10/4)/2 = (5/2)/2 = 5/4 = 1.25

Second root: (-3/4 -13/4)/2 = (-16/4)/2 = (-4)/2 = -2

So y = 1.25 and y = -2.

Check for y = -2: equation x + 1/x = -2. Multiply by x: x² + 2x + 1 = 0 ⇒ (x +1)^2 = 0 ⇒ x = -1. Therefore, x = -1 is a root.

Indeed, plugging x = -1 into original equation: (-1)^4 + (3/4)(-1)^3 - 0.5 (-1)^2 + (3/4)(-1) + 1 = 1 - 3/4 - 0.5 - 3/4 + 1 = 1 - 0.75 - 0.5 - 0.75 + 1 = (1 + 1) - (0.75 + 0.5 + 0.75) = 2 - 2 = 0. Therefore, x = -1 is a root. Therefore, this point is in S.

Therefore, our analysis holds.

Therefore, the area of S is 1/4.

Wait, but the problem says "the set of points (a,b) with 0 ≤ a, b ≤ 1 such that the equation has at least one real root". However, according to our analysis, the area is 1/4, but I want to make sure there isn't something missing.

Wait, could there be cases where even though b < 2 - 2a, the original equation still has a real root?

Let me take a point not in S, say a = 3/4, b = 0.4. Since 2 - 2*(3/4) = 0.5, and 0.4 < 0.5, so this point is not in S. Let me check if the equation has real roots.

Original equation: x⁴ + (3/4)x³ - 0.4x² + (3/4)x + 1 = 0.

Again, using substitution. Divide by x²:

x² + (3/4)x - 0.4 + (3/4)/x + 1/x² = 0

Let y = x + 1/x. Then x² + 1/x² = y² - 2.

Equation becomes: y² - 2 + (3/4)y - 0.4 = 0 ⇒ y² + (3/4)y - 2.4 = 0

Solve for y: y = [ -3/4 ± sqrt( (3/4)^2 + 9.6 ) ] / 2

Compute discriminant: (9/16) + 9.6 = 9/16 + 153.6/16 = 162.6/16 ≈ 10.1625

sqrt(10.1625) ≈ 3.19

Therefore, y ≈ [ -0.75 ± 3.19 ] / 2

First root: (-0.75 + 3.19)/2 ≈ 2.44 / 2 ≈ 1.22

Second root: (-0.75 - 3.19)/2 ≈ -3.94 / 2 ≈ -1.97

Check for real x:

For y ≈ 1.22: equation x + 1/x = 1.22. Multiply by x: x² - 1.22x + 1 = 0. Discriminant ≈ 1.4884 - 4 ≈ -2.5116 < 0. No real roots.

For y ≈ -1.97: equation x + 1/x = -1.97. Multiply by x: x² + 1.97x + 1 = 0. Discriminant ≈ (1.97)^2 - 4 ≈ 3.88 - 4 ≈ -0.12 < 0. Therefore, no real roots. Hence, no real roots for this point.

Therefore, the area calculation seems correct.

Therefore, the answer is 1/4.

But wait, I remember that sometimes when you do these substitutions, you might miss some roots? Let me think again.

We started with the quartic equation, divided by x², substituted y = x + 1/x, and transformed into a quadratic in y. Then, we determined that for the quartic to have a real root, the quadratic must have a root with |y| ≥ 2.

But suppose that x = 1 or x = -1? Let me check if these can be roots.

For example, if x = 1: Plug into the original equation: 1 + a - b + a + 1 = 2 + 2a - b. So 2 + 2a - b = 0. Therefore, if 2 + 2a - b = 0, then x = 1 is a root. Similarly, x = -1: (-1)^4 + a(-1)^3 - b(-1)^2 + a(-1) + 1 = 1 -a - b -a + 1 = 2 - 2a - b. Therefore, if 2 - 2a - b = 0, then x = -1 is a root.

Wait, so these are specific cases. But in our previous analysis, we considered that the quartic has a real root when either f(2) ≤ 0 or f(-2) ≤ 0. Which translates to 2 + 2a - b ≤ 0 or 2 - 2a - b ≤ 0.

But if 2 + 2a - b = 0, then x = 1 is a root, so our condition includes that case. Similarly, 2 - 2a - b = 0 implies x = -1 is a root.

Therefore, our analysis includes those cases.

But in our region S, we only considered the condition b ≥ 2 - 2a. But when b = 2 - 2a, then 2 - 2a - b = 0, which is exactly when x = -1 is a root. Therefore, the boundary of our region S is exactly when x = -1 is a root.

Therefore, our region S includes exactly those points (a, b) where the original equation has a real root. Therefore, the area is indeed 1/4.

But let me check if there are other real roots which are not captured by this analysis. For example, suppose there's a real root x ≠ ±1. But according to our substitution, any real root x ≠ 0 would lead to a solution y with |y| ≥ 2. Therefore, our analysis should capture all possibilities.

Therefore, the answer should be 1/4.

But let me check once more with another example. Let me take a = 1/2, b = 1. As we saw earlier, the equation has a root at x = -1. Let me check. Plug x = -1 into the equation: (-1)^4 + (1/2)(-1)^3 - 1*(-1)^2 + (1/2)(-1) + 1 = 1 - 1/2 -1 -1/2 +1 = 1 - 0.5 -1 -0.5 +1 = (1 + 1) - (0.5 + 1 + 0.5) = 2 - 2 = 0. Yes, it works.

Another case: a = 1, b = 1/2. Then according to the region, since a = 1 ≥ 1/2, and b = 1/2 ≥ 2 - 2*1 = 0. So yes. Let me check if the equation has a real root.

Original equation: x⁴ + x³ - (1/2)x² + x + 1 = 0.

Again, using substitution. Divide by x²:

x² + x - 1/2 + 1/x + 1/x² = 0

Let y = x + 1/x, so x² + 1/x² = y² - 2. Then equation becomes:

y² - 2 + x - 1/2 + 1/x = 0. Wait, no, original equation:

Wait, the original equation after dividing by x² is x² + x - 1/2 + 1/x + 1/x² = 0. Which is (x² + 1/x²) + x + 1/x - 1/2 = 0.

So that is (y² - 2) + y - 1/2 = 0 ⇒ y² + y - 2.5 = 0.

Solving: y = [-1 ± sqrt(1 + 10)] / 2 = [-1 ± sqrt(11)] / 2 ≈ [-1 ± 3.3166]/2.

First root: (-1 + 3.3166)/2 ≈ 2.3166 / 2 ≈ 1.1583

Second root: (-1 - 3.3166)/2 ≈ -4.3166 / 2 ≈ -2.1583

Check for the second root, y ≈ -2.1583. Then equation x + 1/x = -2.1583. Multiply by x: x² + 2.1583x + 1 = 0. Discriminant ≈ (2.1583)^2 - 4 ≈ 4.658 - 4 = 0.658 > 0. Therefore, real roots. Therefore, there is a real root.

Hence, the point (a=1, b=1/2) is in S, which is correct.

Therefore, after thorough checking, I believe that the area of S is indeed 1/4.

But just to be thorough, let me think if there's another approach. For example, suppose we consider the original quartic equation.

We can analyze the function f(x) = x⁴ + a x³ - b x² + a x + 1. We want to know for which (a,b) there exists x ∈ ℝ such that f(x) = 0.

Since f(x) is a continuous function, we can consider its minimum value. If the minimum value of f(x) is ≤ 0, then there's a real root.

But finding the minimum of a quartic is complicated. However, given the symmetry we found earlier, maybe we can use it.

Alternatively, considering the substitution we did earlier, which reduced the problem to analyzing y.

But our previous approach seems valid, and all test cases confirm that.

Therefore, I think confidently that the area is 1/4.

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

To solve the problem, we are given a quartic polynomial:

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

and asked to determine the **area** of the region $ S $ in the unit square $ 0 \leq a, b \leq 1 $ such that this polynomial has **at least one real root**. The goal is to find the set $ S $ and then compute its area.

---

### Step 1: Analyze the Equation

We make the substitution $ y = x + \frac{1}{x} $. This is useful because the polynomial is **symmetric** around the middle term, and dividing through by $ x^2 $ (valid when $ x \neq 0 $) gives:

$$
x^4 + ax^3 - bx^2 + ax + 1 = 0 \Rightarrow x^2 + a x - b + \frac{a}{x} + \frac{1}{x^2} = 0
$$

Rewriting in terms of $ y = x + \frac{1}{x} $, we can use:
$$
x^2 + \frac{1}{x^2} = y^2 - 2
$$
So the equation becomes:
$$
y^2 - 2 + a y - b = 0 \Rightarrow y^2 + a y - (b + 2) = 0
$$

This is a **quadratic in $ y $**. For the original equation to have any **real roots** in $ x $, some real solution $ y $ must be such that $ x + \frac{1}{x} = y $ has a **real solution** $ x $.

This is guaranteed to have real solutions $ x \neq 0 $ **if and only if**:

$$
|y| \geq 2
$$

---

### Step 2: Characterizing the Set $ S $

We now find the set of $ (a, b) $ such that the quadratic equation $ y^2 + a y - (b + 2) = 0 $ has **at least one real root** with $ |y| \geq 2 $.

Let’s denote the two roots of the quadratic as $ y_1 $ and $ y_2 $. Since the quadratic has real roots, it is always true (as $ a, b \in [0,1] \Rightarrow a^2 + 4(b+2) > 0 $), and we only need to check if **at least one root** satisfies $ |y| \geq 2 $.

We use the **test values** at $ y = 2 $ and $ y = -2 $, since the sign of the quadratic function at these points can characterize the root locations.

Let:

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

Then:

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

We consider cases where there is at least one real root with $ |y| \geq 2 $, which is guaranteed if:

$$
f(2) \leq 0 \quad \text{or} \quad f(-2) \leq 0
$$

We are to find the region in $ 0 \leq a, b \leq 1 $ such that at least one of these is true.

---

### Step 3: Determine the Region $ S $

We analyze:

1. $ f(2) \leq 0 \Rightarrow 2 + 2a - b \leq 0 \Rightarrow b \geq 2 + 2a $

   Since $ a \in [0, 1] $, $ 2 + 2a \in [2, 4] $, and $ b \in [0, 1] $, this condition **cannot ever** hold.

2. $ f(-2) \leq 0 \Rightarrow 2 - 2a - b \leq 0 \Rightarrow b \geq 2 - 2a $

This condition defines a **region** in the unit square $ [0,1] \times [0,1] $, and since $ a \in [0,1] $, we find:

- $ 2 - 2a \in [0, 2] $, so $ b \geq 2 - 2a $ is possible only when $ b \in [2 - 2a, 1] $, as $ b \leq 1 $

Thus, the set $ S $ is defined by:

$$
S = \left\{ (a, b) \in [0,1] \times [0,1] \mid b \geq 2 - 2a \right\}
$$

This is a **triangle** in the $ a $-$ b $ plane, bounded by:

- $ b = 2 - 2a $, the line from $ (a, b) = (1, 0) $ to $ (1/2, 1) $
- $ a = 1/2 $, where $ b = 1 $
- $ a = 1 $, where $ b = 0 $
- and the vertical lines $ a = 0 $ and $ b = 0 $, but only the triangle defined above is included in the region.

This triangle has:

- Base from $ a = 1/2 $ to $ a = 1 $, length $ 1/2 $
- Height from $ b = 0 $ to $ b = 1 $, length $ 1 $

So the area is:

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

---

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