The quadratic equation $3x^2+4x-9 = 2x^2-6x+1$ has two real roots. What is the sum of the squares of these roots?
+129883 3x^2+4x-9 = 2x^2-6x+1 rearrange as
x^2 + 10x - 10 = 0
By Vieta
Call the roots a and b
The sum of the roots = -10/1 = -10
So
a +b = -10 squsre both sides
a^2 + 2ab + b^2 =100 (1)
And the product of the roots = -10/1 = -10
So
ab = -10 and 2ab = -20 (2)
Sub (2) into (1)
a^2 - 20 + b^2 = 100
a^2 + b^2 = 120 = the sum of the squares of the roots
