Find the number of ordered pairs (a,b) of real numbers such that
\(a\) is a root of \(x^2+ax+b\)
\(b\) is a root of \(x^2+ax+b\)
If a is a root, then
a^2 + a(a) + b = 0
2a^2 + b = 0
b = -2a^2 (1)
If b is a root, then
b^2 + ab + b = 0 (2)
Sub (1) into (2)
(-2a^2)^2 +a(-2a^2) + (-2a^2) = 0 simplify
4a^4 -2a^3 - 2a^2 = 0 factor
(2a^2) (2a^2 - a - 1) = 0
Set both factors to 0 and solve
2a^2 = 0 (2a + 1) (a - 1) = 0
a = 0 a = -1/2 a = 1
Using (1).....
When a = 0, b = 0 (trivial solution)
When a = -1/2 , b = -1/2
When a = 1, b = -2
Proof
x^2 + 0x +0 = 0 .......has the root x = 0.....(a,b) = (0,0)
x^2 - (1/2)x - 1/2 = 0 multiply through by 2
2x^2 - x^2 - 1 = 0 factor
(2x + 1) (x - 1) = 0
Has the roots x = -1/2 and x = 1
So .....(a,b) = (-1/2, -1/2)
x^2 -2x + 1 = 0 factor
(x -1) (x -1) = 0.......so x = 1 is a root......(a, b) = (1, -2)
So......
(a, b) = (0, 0) (trivial)
(a, b) = (-1/2, -1/2)
(a, b) = (1, -2)