Let u and v be the solutions to 3x^2 + 5x + 7 = 2x^2 + 10x + 4. Find 1/u^2 + 1/v^2.
+289 x^2 - 5x + 3 = 0
The polynomial that has the recirpocal of the roots of the equation above is:
3x^2 - 5x + 1 = 0
We just wnat the polynomial with the square of its roots.
Suppose, that polynomial above is P(x) and has roots a and b. And let Q(x) be the polynomial that has the square of its roots.
P(x) = (x - a)(x - b)
Q(x) = (x - a^2)(x - b^2)
Q(x^2) = (x-a)(x-b) * (x + a)(x + b)
= P(x)P(-x)
= (3x^2 - 5x + 1)(3x^2 + 5x + 1)
= (3x^2 + 1)^2 - 25x^2
= 9x^4 + 6x^2 + 1 - 25x^2
= 9x^4 - 19x^2 + 1
Q(x) = 9x^2 - 19x + 1
The sum of the roots of this equation is 19/9
