Question

solve the above question​

Answer 1

Step-by-step explanation:

Given expression,

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

Rationalizing it,

$\sf \bf :\longmapsto x = \dfrac{\sqrt{p+2q} + \sqrt{p - 2q}}{\sqrt{p+2q}- \sqrt{p-2q}} \times \dfrac{\sqrt{p+2q} + \sqrt{p - 2q}}{\sqrt{p+2q}+ \sqrt{p-2q}}$

$\sf \bf :\longmapsto x = \dfrac{(\sqrt{p+2q} + \sqrt{p - 2q})^2}{(\sqrt{p+2q})^2- (\sqrt{p-2q})^2}$

$\sf \bf :\longmapsto x = \dfrac{p+2q +p - 2q +2+\sqrt{(p+2q)(p-2q)}}{p+2q -p+2q}$

$\sf \bf :\longmapsto x= \dfrac{2p + 2\sqrt{p^2-4q^2}}{4q}$

$\sf \bf :\longmapsto x=\dfrac{2(p+\sqrt{p^2-4q^2}}{4q}$

$\sf \bf :\longmapsto x=\dfrac{p+\sqrt{p^2 - 4q^2}}{2q}$

$\sf \bf :\longmapsto 2qx = p + \sqrt{p^2-4q^2}$

$\sf \bf :\longmapsto 2qx - p = \sqrt{p^2 - 4q^2}$

Squaring both sides,

$\sf \bf :\longmapsto 4q^2x^2 + p^2 - 4pqx = p^2 - 4q^2$

$\sf \bf :\longmapsto 4q^2x^2 -4pqx = -4q^2$

$\sf \bf :\longmapsto pqx - q^2x^2 = q^2$

$\sf \bf :\longmapsto q(px - qx^2)=q^2$

$\sf \bf :\longmapsto px - qx^2 = q^2$

$\sf \bf :\longmapsto qx^2 - px + q^2 = 0$

Hence ,

$\boxed{\boxed{\sf \bf qx^2 - px + q^2 = 0}}$

Additional information:

Identities used:

$\large \boxed{\begin{array}{l} \sf \bf (a+b)^2 = a^2 + 2ab + b^2 \\ \sf \bf (a+b)(a-b) = a^2-b^2\\ \sf \bf (a-b)^2 = a^2 - 2ab + b^2\end{array}}$