Question

If, x = [root(p+q) +root(p-q)] /[root(p+q) - root(p-q)] then find the value of qx2 – 2px + q .

Answer 1

The correct answer is,  

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

Answer 2

Answer:

qx² - 2px + q = 0

Step-by-step explanation:

If, x = [root(p+q) +root(p-q)] /[root(p+q) - root(p-q)] then find the value of qx2 – 2px + q .

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

Multiplying & diving by √(p + q)   + √(p-q)

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

=> x  = 2 ( p + √(p² - q²) ) / 2q

=> x = ( p + √(p² - q²) ) / q

=> qx = p + √(p² - q²)

=> qx - p = √(p² - q²)

Squaring both sides

=> q²x² +  p² - 2pqx  = p² - q²

=> q²x² - 2pqx  = - q²

diving by q both sides

=> qx² - 2px = -q

=> qx² - 2px + q = 0