Math, asked by seemasssingh4p893y4, 1 year ago

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Answered by deeksha912
0

Given x/1 = √p+2q + √p-2q / √p+2q - √p-2q

By using componendo and dividendo

⇒ x+1/ x-1 = (√p+2q + √p-2q) + (√p+2q - √p-2q) / (√p+2q + √p-2q) - (√p+2q - √p-21)

⇒ x+1/x-1 = √p+2q + √p-2q +√p+2q-√p-2q / √p+2q + √p-2q-√p+2q+√p-2q

⇒ x+1/x-1 = √p+2q +√p+2q/ √p-2q + √p-2q

⇒ x+1/x-1 = 2√p+2q/2√p-2q

⇒ squaring both side

⇒ (x+1)²/(x-1)² = (√p+2q)²/(p-2q)²

⇒x²+1+2x / x²+1-2x = p+2q /p-2q

⇒(p-2q)(x²+1+2x)=(p+2q)(x²+1-2x)

⇒px²+p+2px-2qx²-2q-4qx = px²+p-2px+2qx²+2q-4qx

⇒2px-2qx²-2q = -2px+2qx²+2q

⇒2px+2px-2qx²-2qx²-2q-2q

⇒4px-4qx²-4q

⇒-4qx²+4px-4q

⇒-4(qx²-px+q)=0

Answer = qx²-px+q=0.

Hope it will help you

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