Question

plz answer it fast it's very urgent​

Answer 1

Sol: Given x = [√(p + 2q) + √(p - 2q) ] / [√(p + 2q) - √(p - 2q)] rationalise the denominator x = [√(p + 2q) + √(p - 2q) ]2 / [(p + 2q) - (p - 2q)] x = [√(p + 2q) + √(p - 2q) ]2 / 4q. 4qx = 2p + 2√(p2 - 4q2) 2qx - p = √(p2 - 4q2) squaring on both sides 4q2x2 - 4pqx + p2 = p2 - 4q2 4q[ qx2 - px + q ] = 0 qx2 - px + q = 0

Hope it is helpful for you

thanks bro

Answer 2

Answer:

Step-by-step explanation:

Given :

If $x = \frac{\sqrt{p + 2q} + \sqrt{p - 2q}}{\sqrt{p + 2q} - \sqrt{p - 2q}}$

then, Prove : $qx^2 - px + q = 0$

Proof :

$x = \frac{\sqrt{p + 2q} + \sqrt{p - 2q}}{\sqrt{p + 2q} - \sqrt{p - 2q}}$

By rationalising the denominator, we get :

⇒ $x = (\frac{\sqrt{p + 2q} + \sqrt{p - 2q}}{\sqrt{p + 2q} - \sqrt{p - 2q}})(\frac{\sqrt{p + 2q} + \sqrt{p - 2q}}{\sqrt{p + 2q} + \sqrt{p - 2q}})$

⇒ $x = \frac{(\sqrt{p + 2q} + \sqrt{p - 2q})^2}{(\sqrt{p + 2q})^2 - (\sqrt{p - 2q})^2}$

⇒ $\frac{(\sqrt{p + 2q})^2 + (\sqrt{p - 2q})^2 + 2(\sqrt{p + 2q})(\sqrt{p - 2q})^2}{(p + 2q) - (p - 2q)}$

⇒ $\frac{(p + 2q) + ( p - 2q) + 2\sqrt{(p + 2q)(p - 2q)}}{4q}$

⇒ $\frac{2p + 2\sqrt{p^2 - 4q^2} }{4q} =\frac{p + \sqrt{p^2 - 4q^2} }{2q}$

Which is of the form :

⇒ $x = \frac{-b+\sqrt{b^2 - 4ac} }{2a}$

for which the equation is : $ax^2 + bx + c = 0$

⇒ $x = \frac{p + \sqrt{p^2 - 4q^2} }{2q}$

a = c = q , b = -p

THHEN THE EQUATION SHOULD BE : $qx^2 - px + q = 0$

(or)

By substituting the rationalised value of x in the given equation , we get,

⇒ $LHS = qx^2 - px + q$

⇒ $q(\frac{p+\sqrt{p^2 - 4q^2} }{2q})^2 - p(\frac{p+\sqrt{p^2 - 4q^2} }{2q}) + q$

⇒ $q(\frac{p^2 + (p^2 - 4q^2) + 2p(\sqrt{p^2 - 4q^2})}{4q^2}) - p(\frac{p+\sqrt{p^2 - 4q^2} }{2q} ) + q$

⇒ $\frac{2p^2 - 4q^2 + 2p(\sqrt{p^2 - 4q^2)}}{4q} - \frac{p^2+p\sqrt{p^2 - 4q^2} }{2q} + q$

⇒ $\frac{p^2 - 2q^2 + p(\sqrt{p^2 - 4q^2)}}{2q} - \frac{p^2+p\sqrt{p^2 - 4q^2} }{2q} + q$

⇒ $\frac{p^2 - 2q^2 + p(\sqrt{p^2 - 4q^2})-(p^2+p\sqrt{p^2 - 4q^2})}{2q} + q$

⇒ $\frac{- 2q^2}{2q} + q = q - q = 0 = RHS$