Let S be the set of points \((a,b)\) with \(0 \le a,b \le 1\) such that the equation \(x^4 + ax^3 - bx^2 + ax + 1 = 0\)
has at least one real root. Determine the area of the graph of S
As long as b >= 2 - 2a the equation has at least one real root. the line b = 2 - 2a is a straight line going from b = 2 when a = 0, to b = 0 when a = 1. Hence the area of b >= 2 - 2a is a triangle of base 1 and height 2; i.e. the area is (1/2)*1*2 = 1.
I solved the quadratic in terms of a and b (well, ok, I used a piece of software to do so!). This resulted in an expression under a square root sign that meant if b>=2-2a, two of the four roots would be real.
How can the height be 2?, doesnt\(0 \le a,b \le 1\)mean \(0 \le a \le 1\)and \(0 \le b \le 1\)? So it would have to be inside a 1 by 1 square?
Also, I got b<=-2a+2, can you check yours
EDIT: Nevermind, found my error