Question

If x=root p+2q + root p-2q/root p+2q - root p-2q then show that qx square - px + q =0

Answer 1

ANSWER:—

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)+√(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

∴, x²={p+√(p²-4q²)}²/4q²

=[p²+2p√(p²-4q²)+{√(p²-4q²)}²]/4q²

={p²+p²-4q²+2p√(p²-4q²)}/4q²

=2{p²-2q²+p√(p²-4q²)}/4q²

={p²-2q²+p√(p²-4q²)}/2q²

∴, qx²-px+q

=q[{p²-2q²+p√(p²-4q²)}/2q²]-p[{p+√(p²-4q²)}/2q]+q

={p²-2q²+p√(p²-4q²)}/2q-{p²+p√(p²-4q²)}/2q+q

={p²-2q²+p√(p²-4q²)-p²-p√(p²-4q²)+2q²}/2q

=0 (Proved)

hope it helps

thanks

Answer 2

Answer:

In attachment I have answer this question..