Question

x=√p+2q +√p-2q ÷(√p+2q- √p-2q) and q≠0,then prove that ,qx²-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)+√(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)

Answer 2

Answer:

In attachment I have answered this problem↑↑↑↑