Question

If x= root p+2q + root p-2q/ root p+2q - root p-2q , then show that qx square - px +q =0

Answer 1

$x=\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}\\\\rationalize\\\\x=\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}*\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}+\sqrt{p-2q}}\\\\=\frac{p+2q+p-2q+2\sqrt{p^2-4q^2}}{p+2q-(p-2q)}\\\\=\frac{p+\sqrt{p^2-4q^2}}{2q}$

Clearly we can see that  x is a root of  quadratic : q x² - p x + q

Hence,  q x² - p x + q = 0

Answer 2

Aswer::::::::

answer given by Kvnmurti is correct so you can go through that answer.

Step-by-step explanation:

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