Question

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$x = \frac{ \sqrt{p + 2q} + \sqrt{p - 2q} }{ \sqrt{p + 2q} - \sqrt{p - 2q} }$

Then , show that :

qx² - px + q = 0

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@Sanskriti141

Answer 1

$x = \frac{\sqrt{p + 2q} + \sqrt{p - 2q}}{\sqrt{p + 2q} - \sqrt{p - 2q}}$

Applying Componendo & Dividendo Theoram on Both sides :

⇒ $\frac{x + 1}{x - 1} = \frac{\sqrt{p + 2q} + \sqrt{p - 2q} + \sqrt{p + 2q} - \sqrt{p - 2q}}{\sqrt{p + 2q} + \sqrt{p - 2q} - \sqrt{p + 2q} + \sqrt{p - 2q}}$

⇒ $\frac{x + 1}{x - 1} = \frac{2\sqrt{p + 2q}} {2\sqrt{p - 2q}}$

Squaring on both sides We get :

⇒ $\frac{(x + 1)^2}{(x - 1)^2} = \frac{p + 2q}{p - 2q}$

Applying Componendo - Dividendo Theoram again on both sides :

⇒ $\frac{(x + 1)^2 + (x - 1)^2}{(x + 1)^2 - (x - 1)^2} = \frac{p + 2q + p - 2q}{p + 2q - p + 2q}$

⇒ $\frac{2x^2 + 2}{4x} = \frac{2p}{4q}$

⇒ $\frac{x^2 + 1}{x} = \frac{p}{q}$

⇒ qx² - px + q = 0

Answer 2

First of all rationalize the value of x given


After rationalising, cancel the common factors

After obtaining the final value of x, find the value of x²

Again cancel the common factors of x² and you will obtain the final value of x²

Then substitute the value of x² and x in the equation

qx² - px + q


Then you will be able to solve.


Refer the attachments.

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Hope it helps dear friend ☺️✌️