Okay, so I need to find the area of the set S, which consists of points (a, b) where the equation x⁴ + a x³ - b x² + a x + 1 = 0 has at least one real root. The variables a and b are between 0 and 1. Hmm, let's start by understanding the problem step by step.

First, the equation is a quartic (degree 4) polynomial. Quartic equations can be tricky, but maybe there's a way to factor it or analyze its behavior. Since we're looking for real roots, perhaps we can use the Intermediate Value Theorem or Descartes' Rule of Signs. Alternatively, maybe substituting y = x + 1/x or something like that could simplify the equation? Let me think.

Wait, looking at the equation: x⁴ + a x³ - b x² + a x + 1 = 0. I notice that the coefficients are symmetric in a way. The coefficients of x⁴ and the constant term are both 1, the coefficients of x³ and x are both a, and the middle term is -b x². Maybe this symmetry can be exploited.

Let me try substituting y = x + 1/x. If I let y = x + 1/x, then y² = x² + 2 + 1/x², and y³ = x³ + 3x + 3/x + 1/x³. Hmm, but our equation has x⁴, x³, x², x, and constants. If I multiply both sides by x², maybe?

Multiplying the equation by x² gives: x⁶ + a x⁵ - b x⁴ + a x³ + x² = 0. That's a sixth-degree polynomial. Hmm, not sure if that's helpful. Alternatively, maybe dividing by x² to make it symmetric?

Divide the original equation by x²: x² + a x - b + a/x + 1/x² = 0. Then, group terms: (x² + 1/x²) + a(x + 1/x) - b = 0. Let me let y = x + 1/x again. Then, x² + 1/x² = y² - 2. So substituting, we get (y² - 2) + a y - b = 0. So the equation becomes y² + a y - (b + 2) = 0.

That's a quadratic in y! That seems promising. So if we let y = x + 1/x, then the equation reduces to y² + a y - (b + 2) = 0. Now, for real roots x, y must satisfy y = x + 1/x. But x can be positive or negative. Wait, but since a and b are between 0 and 1, and x is a real number. Let me think about the possible values of y.

If x > 0, then y = x + 1/x ≥ 2 by AM ≥ GM. If x < 0, let's say x = -t where t > 0, then y = -t - 1/t = -(t + 1/t) ≤ -2. So y can be ≥ 2 or ≤ -2. So the equation y² + a y - (b + 2) = 0 must have a solution y with |y| ≥ 2. Therefore, the original quartic equation has a real root if and only if the quadratic equation in y has a real solution y with |y| ≥ 2.

Therefore, we can rephrase the problem: find all (a, b) in [0,1]² such that the quadratic equation y² + a y - (b + 2) = 0 has a real solution y with |y| ≥ 2. Then, the area of S is the area of such (a, b) pairs.

So, let's analyze the quadratic equation y² + a y - (b + 2) = 0. For this quadratic to have real solutions, the discriminant must be non-negative. The discriminant D = a² + 4(b + 2) ≥ 0. But since a and b are non-negative, D is always positive. So the quadratic always has two real roots. However, we need at least one of these roots to satisfy |y| ≥ 2.

So, we need to find (a, b) such that at least one root of y² + a y - (b + 2) = 0 is ≥ 2 or ≤ -2.

Let me denote the roots as y1 and y2. Then, since the quadratic is y² + a y - (b + 2) = 0, the roots are y = [-a ± sqrt(a² + 4(b + 2))]/2.

We need either y1 ≥ 2 or y2 ≤ -2.

But since the product of the roots y1 y2 = - (b + 2). So if one root is large positive, the other must be negative with a large magnitude. So, if y1 ≥ 2, then y2 = (- (b + 2))/y1 ≤ - (b + 2)/2. Similarly, if y1 ≤ -2, then y2 = (- (b + 2))/y1 ≥ (b + 2)/2. But since the quadratic can have two roots, maybe both roots are on one side? Wait, but the quadratic y² + a y - (b + 2) = 0. Let's consider the quadratic function f(y) = y² + a y - (b + 2). The parabola opens upwards. The roots are where f(y) = 0.

So, to have a root y ≥ 2, the quadratic must cross y = 2, so f(2) ≤ 0. Similarly, to have a root y ≤ -2, the quadratic must cross y = -2, so f(-2) ≤ 0.

Therefore, the conditions are:

Either f(2) ≤ 0 or f(-2) ≤ 0.

Let me compute f(2):

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

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

So, f(2) ≤ 0 implies 2a - b + 2 ≤ 0, which is equivalent to b ≥ 2a + 2.

But since a ≥ 0 and b ≤ 1, 2a + 2 ≥ 2*0 + 2 = 2. However, b ≤ 1, so b ≥ 2a + 2 is impossible because 2a + 2 ≥ 2 > 1 ≥ b. Therefore, f(2) ≤ 0 is impossible.

Similarly, f(-2) ≤ 0 implies -2a - b + 2 ≤ 0, which is equivalent to 2a + b ≥ 2. So, the condition is 2a + b ≥ 2.

But since a and b are between 0 and 1, 2a + b can be at most 2*1 + 1 = 3, which is more than 2. So the region where 2a + b ≥ 2 is a subset of [0,1]².

Therefore, the set S is the set of all (a, b) in [0,1]² such that 2a + b ≥ 2. Therefore, the area of S is the area of the region where 2a + b ≥ 2 in the unit square [0,1]².

But wait, let me check if this is correct. Because we concluded that f(2) ≤ 0 is impossible, so the only way the quadratic has a root with |y| ≥ 2 is if there's a root y ≤ -2, which requires f(-2) ≤ 0, i.e., 2a + b ≥ 2.

But we need to verify if this is sufficient. Suppose 2a + b ≥ 2, then the quadratic has a root ≤ -2, which in turn implies that the original quartic equation has a real root. Conversely, if the quartic equation has a real root, then there exists a y with |y| ≥ 2, which implies that either f(2) ≤ 0 or f(-2) ≤ 0. But f(2) ≤ 0 is impossible, so it must be that f(-2) ≤ 0, hence 2a + b ≥ 2.

Therefore, the set S is exactly the set of (a, b) in [0,1]² with 2a + b ≥ 2. Therefore, the area is the area of the region in the unit square above the line 2a + b = 2.

Let me plot this line in the unit square. The line 2a + b = 2. When a = 0, b = 2, which is outside the unit square. When b = 0, a = 1. So the line intersects the unit square at (1, 0). But wait, since a and b are between 0 and 1, so when a = 1, 2*1 + b = 2 + b. For this to equal 2, b must be 0. So the line passes through (1, 0) and, when b = 1, 2a + 1 = 2 => a = 0.5. So another point is (0.5, 1). So the line connects (0.5, 1) to (1, 0). Therefore, the region 2a + b ≥ 2 in the unit square is the triangle with vertices at (0.5, 1), (1, 1), and (1, 0).

Wait, let me check. If we have 2a + b = 2, then for a between 0.5 and 1, b goes from 1 down to 0. So the region above the line is the triangle bounded by (0.5, 1), (1, 1), and (1, 0). Therefore, the area is the area of this triangle.

The base of the triangle is from (0.5, 1) to (1, 1), which is length 0.5. The height is from (1, 1) down to (1, 0), which is 1. So the area would be (base * height)/2 = (0.5 * 1)/2 = 0.25. But wait, that seems too small. Let me verify.

Alternatively, the area can be calculated as the integral over a from 0.5 to 1 of (2 - 2a) da, since b must be ≥ 2 - 2a. So the area is ∫ from a=0.5 to a=1 of (2 - 2a) da.

Calculating that integral:

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

At a=1: 2*1 - 1² = 2 - 1 = 1.

At a=0.5: 2*0.5 - (0.5)² = 1 - 0.25 = 0.75.

So the integral is 1 - 0.75 = 0.25. So the area is 0.25. But wait, that seems correct? Because the region is a triangle with base 0.5 and height 1, so area is 0.25. But let me confirm.

Alternatively, the area can be calculated as the area of the unit square minus the area below the line 2a + b = 2. The area below the line is a polygon with vertices at (0,0), (0,2), but since b is limited to 1, the intersection point is at (0.5, 1). So the area below the line in the unit square is the area of the polygon from (0,0) to (0,1) to (0.5,1) to (1,0). Wait, that's a quadrilateral. Hmm, maybe it's better to split the area.

Wait, actually, the line 2a + b = 2 intersects the unit square at (1, 0) and (0.5, 1). So the area below the line is the triangle formed by (0,0), (0,1), (0.5,1), and (1,0). Wait, no. Wait, the line goes from (0.5, 1) to (1, 0). So the area below the line is the quadrilateral with vertices (0,0), (0,1), (0.5,1), (1,0). To find the area of this quadrilateral, we can split it into two parts: the rectangle from (0,0) to (0,1) to (0.5,1) to (0.5,0.5) to (1,0)? Wait, maybe not. Alternatively, we can use the formula for the area of a polygon given coordinates.

Alternatively, the area below the line can be found as the sum of two areas: the rectangle from a=0 to a=0.5, b=0 to b=1, which has area 0.5*1=0.5, and the triangle from a=0.5 to a=1, b=1 down to b=0. The area of that triangle is 0.5*0.5*1=0.25. So total area below the line is 0.5 + 0.25 = 0.75. Therefore, the area above the line is 1 - 0.75 = 0.25. So that's consistent.

Therefore, the area of S is 0.25. But wait, let me check again. If the line is from (0.5,1) to (1,0), then the area above the line is the triangle with vertices (0.5,1), (1,1), and (1,0). The base of this triangle is from (0.5,1) to (1,1), which is 0.5 units. The height is from (1,1) down to (1,0), which is 1 unit. The area of a triangle is (base * height)/2 = (0.5 * 1)/2 = 0.25. So yes, 0.25.

But let me ensure that this is indeed the correct region. Suppose a=0.5, b=1. Then 2a + b = 2*0.5 + 1 = 2, so the point (0.5,1) is on the line. If a=1, b=0, then 2*1 + 0 = 2, so that's also on the line. If a=0.75, then b must be 2 - 2*0.75 = 2 - 1.5 = 0.5. So the line goes through (0.75, 0.5), which is in the unit square. So indeed, the region above the line is a triangle with vertices at (0.5,1), (1,1), (1,0), and (0.5,1) again? Wait, no. Wait, the region above the line 2a + b = 2 in the unit square is the set of points (a, b) where 2a + b ≥ 2. Since 2a + b is a linear function, the region is a convex polygon. The intersection points of the line 2a + b = 2 with the unit square are (0.5, 1) and (1, 0). Therefore, the region above the line is the polygon bounded by (0.5,1), (1,1), (1,0), and back to (0.5,1). Wait, but (1,1) is not on the line. Wait, no. The region 2a + b ≥ 2 in the unit square is the area above the line, which would be a triangle with vertices at (0.5,1), (1,1), and (1,0). But (1,1) is not on the line; the line at a=1 is b=0. So the region is a quadrilateral? Wait, I'm getting confused.

Wait, maybe it's better to visualize. The line 2a + b = 2 passes through (0.5,1) and (1,0). In the unit square [0,1]x[0,1], these are two points. The region 2a + b ≥ 2 is the area above this line. So for a between 0.5 and 1, the line defines a boundary where b = 2 - 2a. So above this line, b ≥ 2 - 2a. Since a can be from 0.5 to 1, b must be from 2 - 2a up to 1. For a=0.5, b must be ≥ 1, which is the point (0.5,1). For a=1, b must be ≥ 0, which is the point (1,0). So the region is a polygon bounded by the line from (0.5,1) to (1,0), the top edge of the square from (0.5,1) to (1,1), and the right edge from (1,1) to (1,0). Wait, no, that's not correct. The region above the line is bounded by the line itself, the top edge, and the right edge. Wait, actually, the line goes from (0.5,1) to (1,0), so the region above the line is the area above this line, which is a triangle with vertices at (0.5,1), (1,1), and (1,0). Wait, but (1,1) is not on the line. The line at a=1 has b=0. So the region is a quadrilateral with vertices at (0.5,1), (1,1), (1,0), and (0.5,1). Wait, that can't be. Wait, perhaps the region is a triangle with vertices at (0.5,1), (1,1), and (1,0). Let me check.

Wait, the area where 2a + b ≥ 2 in the unit square is all (a, b) with a in [0.5,1] and b ≥ 2 - 2a. So when a is 0.5, b must be ≥ 1. As a increases to 1, the lower bound on b decreases to 0. So the region is a triangle with vertices at (0.5,1), (1,1), and (1,0). Because for each a from 0.5 to 1, b goes from 2 - 2a up to 1. So the figure is a triangle with those three vertices. The area of that triangle is (base * height)/2. The base along a from 0.5 to 1 is 0.5, and the height along b from 0 to 1 is 1. But actually, the triangle is a right triangle with legs of length 0.5 and 1. Wait, no. The base is 0.5 (from a=0.5 to a=1), and the height is 1 (from b=0 to b=1). The area is (0.5 * 1)/2 = 0.25. So that's consistent. Therefore, the area of S is 0.25.

But let me verify this conclusion once more. Suppose we pick a point (a, b) in the region above the line. For example, (0.75, 0.5): 2*0.75 + 0.5 = 1.5 + 0.5 = 2, so it's on the line. If we pick (0.6, 0.8): 2*0.6 + 0.8 = 1.2 + 0.8 = 2, also on the line. If we pick (0.7, 0.7): 2*0.7 + 0.7 = 1.4 + 0.7 = 2.1 ≥ 2, so it's in the region. If we pick (0.5, 1): 2*0.5 + 1 = 2, on the line. If we pick (1, 0): 2*1 + 0 = 2, on the line. So all points above the line in the unit square satisfy 2a + b ≥ 2. Therefore, the area is indeed 0.25.

Therefore, the area of S is 1/4. So the final answer is \boxed{\dfrac{1}{4}}.

**Final Answer**
The area of the graph of \( S \) is \boxed{\dfrac{1}{4}}.
Therefore, the area of the set \( S \) is \(\boxed{\frac{1}{4}}\).
</think>