Let u and v be the solutions to 3x^2 + 5x + 7 = 2x^2 + 10x + 4. Find 1/u^2 + 1/v^2.

Guest Jul 21, 2022

#1**0 **

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**

Voldemort Jul 22, 2022