if x is equal to root P + Q + root 3 minus Q by root P + Q minus root 3 minus Q then find the value of q x square - 2 PX + q
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Answer:
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
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Step-by-step explanation:
0 is the correct answer..
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