Question

If x =√(p+2q) +√(p-2q) divided by√(p+2q) +√(p-2q), Show that qx 2 - px + q =0

Answer 1

x=√(p+2q)+√(p-2q)/√(p+2q)-√(p-2q)
={√(p+2q)+√(p-2q)}{√(p+2q)+√(p-2q)}/{√(p+2q)-√(p-2q)}{√(p+2q)+√(p-2q)}
=[{√(p+2q)}²+2√(p+2q)√(p-2q)+{√(p-2q)}²]/[{√(p+2q)}²-{√(p-2q)}²]
={p+2q+p-2q+2√(p+2q)(p-2q)}/(p+2q-p+2q)
={2p+2√(p²-4q²)}/4q
={p+√(p²-4q²)}/2q
∴, qx²-px+q
=q[{p+√(p²-4q²)}²/4q²]-p{p+√(p²-4q²)}/2q+q
=p²+2p√(p²-4q²)+p²-4q²}/4q-{p²+p√(p²-4q²)}/2q+q
=2p²+2p√(p²-4q²)-4q²-2p²-2p√(p²-4q²)+4q²
=0 (Proved)

Answer 2

Rationalize the denominator in x.

$x=\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}\\\\=\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}-\sqrt{p-2q}}*\frac{\sqrt{p+2q}+\sqrt{p-2q}}{\sqrt{p+2q}+\sqrt{p-2q}}\\\\=\frac{p+2q+p-2q+2\sqrt{(p+2q)(p-2q)}}{(p+2q)-(p-2q)}\\\\x=\frac{p+\sqrt{p^2-4q^2}}{2q}\\\\\implies\ 2qx-p=\sqrt{p^2-4q^2},\\\\ squaring\ both\ sides\\\\4q^2x^2+p^2-4pqx=p^2-4q^2\\\\Simplifying,\ qx^2-px+q=0$