Question

Prove that qx^2 - px + q = 0If x = √p+2q + √p-2q / √p+2q - √p-2q and q ≠ 0, then prove that qx^2 - px + q =0

Answer 1

Given x = [√(p + 2q) + √(p - 2q) ] / [√(p + 2q) - √(p - 2q)],,
rationalise the denominator,,
x = [√(p + 2q) + √(p - 2q) ]2 / [(p + 2q) - (p - 2q)],,
x = [√(p + 2q) + √(p - 2q) ]2 / 4q.,,
4qx = 2p + 2√(p2 - 4q2),,
2qx - p = √(p2 - 4q2) squaring on both sides,,
4q2x2 - 4pqx + p2 = p2 - 4q2,,

4q[ qx2 - px + q ] = 0,,

qx2 - px + q = 0.

Answer 2

x = [√(p + 2q) + √(p - 2q) ]2 / [(p + 2q) - (p - 2q)],,
x = [√(p + 2q) + √(p - 2q) ]2 / 4q.,,
4qx = 2p + 2√(p2 - 4q2),,
2qx - p = √(p2 - 4q2) 
4q2x2 - 4pqx + p2 = p2 - 4q2,,
4q[ qx2 - px + q ] = 0,,
qx2 - px + q = 0.