0

Let consider a second degree $$a(k)x^2+b(k)x+c(k)=0$$

Suppose I want to find conditions on $k$ so that, called $x_1$ and $x_2$ the two solutions of the equation, they satisfy the following

$$\frac{1}{x_1^2} +\frac{1}{x_2^2}=\frac{1}{4}$$

Could you give hints?

Maryam
  • 1,889

1 Answers1

1

Hint:

We have:

\begin{cases} x_1 +x_2 = -\frac{b(k)}{a(k)} \\ x_1x_2 =\frac{c(k)}{a(k)} \end{cases}

and \begin{align} \frac{1}{4} &= \frac{1}{x_1^2} +\frac{1}{x_2^2}\\ &= \frac{x_1^2 +x_2^2}{x_1^2x_2^2}\\ &= \frac{(x_1 +x_2)^2-2x_1x_2}{x_1^2x_2^2} \end{align}

NN2
  • 15,892