Question

x=rootp+q+rootp-q/rootp+q-p-q​

Answer 1

Answer:

Given :

x = (√p+q + √p-q ) / (√p+q - √p-q )

Solution :

x = (√p+q + √p-q ) / (√p+q - √p-q ) * (√p+q + √p-q ) / (√p+q + √p-q )

by (a+b) (a-b) = a² - b²

we get,

x= (√p+q + √p-q )² / ( p+q - (p-q) )

x = p+q+p-q +2 √p+q √p-q / 2q

x = 2p + 2 √p² - q² / 2q

x = p + √p² - q² / q

Now, let us find the value of ,

qx² - 2px + q = q ( p + √p² - q² / q)² -2p ( p + √p² - q² / q) + q

= q( p + √p² - q² )² / q² - 2p ( p + √p² - q² / q) + q

= ( p + √p² - q² )² - 2p ( p + √p² - q² / q) + q² / q

= p² + p² - q² + 2p √p² - q² -2p² -2p √p² - q² +q² /q

On cancelling the positive and negative terms,

the numerator would be 0.

So, 0 / q = 0

qx² - 2px + q = 0