Question

prove that q x²-2px+ q = 0

Answer 1

Hey there,

Sol:

Given x = [√(p + 2q) + √(p - 2q) ] / [√(p + 2q) - √(p - 2q)]

rationalize 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

Hope this helps!