Question

If the roots of px^2 +qx + q = 0 are alpha and beta then prove that
root alpha/beta + root beta/alpha + root q/p = 0

Answer 1

$\Large{\underline{\underline{\mathfrak{\pink{\bf{ANSWER}}}}}}$

$\LARGE{\underline{\bf{\:Given}}}$

$\:\:\:\:\:\:\sf{\: (Equation)}$

$\bigstar\sf{\:(px^2+qx+q)\:=\:0}$

$\bigstar\sf{\:\alpha\:and\:\beta\:are\:two\:toots)}$

$\LARGE{\underline{\bf{\:To\:Prove}}}$

$\bigstar\sf{\:\left(\sqrt{\dfrac{\alpha}{\beta}}+\sqrt{\dfrac{\beta}{\alpha}}\right)\:=\:-\left(\sqrt{\dfrac{q}{p}}\right)}$

$\LARGE{\underline{\bf{\:Prove}}}$

We Know,

$\green{\boxed{\boxed{\orange{\sf{\:Sum\:Of\:Roots\:=\:\dfrac{-(Coefficient\:of\:x)}{(Coefficient\:Of\:x^2)}}}}}}$

$\mapsto\pink{\sf{\:(\alpha+\beta)\:=\:\dfrac{-q}{p}}}$.....[equ(1)]

Again,

$\green{\boxed{\boxed{\orange{\sf{\:Product\:Of\:Roots\:=\:\dfrac{(Constant\:part)}{(Coefficient\:Of\:x^2)}}}}}}$

$\mapsto\pink{\sf{\:(\alpha\beta)\:=\:\dfrac{q}{p}}}$.....[equ(2)]

Take L.H.S.

$\red{\sf{\:\left(\sqrt{\dfrac{\alpha}{\beta}}+\sqrt{\dfrac{\beta}{\alpha}}\right)}}$

$\mapsto\sf{\:\left(\dfrac{(\sqrt{\alpha})^2+(\sqrt{\beta})^2}{(\sqrt{\alpha\beta})}\right)}$

$\mapsto\sf{\:\dfrac{(\alpha+\beta)}{(\sqrt{\alpha\beta})}}$

$\:\:\:\:\:\:\small\green{\sf{\:( Keep\:Value\:by\:equ(1)\:and\:(2) )}}$

$\mapsto\sf{\:\dfrac{\dfrac{-q}{p}}{\sqrt{\dfrac{q}{p}}}}$

$\mapsto\sf{\:-\left(\sqrt{\dfrac{q}{p}}\right)^2\times\left(\sqrt{\dfrac{p}{q}}\right)}$

$\mapsto\sf{\:-\left(\sqrt{\dfrac{q}{p}}\right)\times\cancel{\left(\sqrt{\dfrac{q}{p}}\right)}\times\cancel{\left(\sqrt{\dfrac{p}{q}}\right)}}$

$\mapsto\pink{\sf{\:-\left(\sqrt{\dfrac{q}{p}}\right)}}$

= R.H.S.

That's Proved.

_____________________