Math, asked by Anonymous, 1 year ago


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Help me out with this one :)​

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Answered by Anonymous
4

ÊLLØ'......!!

\Huge{\mathfrak{\underline{\underline{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)

THÅÑKẞ.....!!

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