Question

if sum of roots of the quadratic equation is 1/x+p + 1/x+q = 1/r is 0,show that the product of roots is-p^+q^/2​

Answer 1

ANSWER:

Question:

If sum of roots of the quadratic equation is 1/x+p + 1/x+q = 1/r is 0, show that the product of roots is -(p² + q²)/2

Given:

$\:\:\:\:\bullet\:\:\:\:\dfrac{1}{x+p}+\dfrac{1}{x+q}=\dfrac{1}{r}\\\\\:\:\:\:\bullet\:\:\:\:\text{Sum of roots = 0}$

To Prove:

$\:\:\:\:\bullet\:\:\:\:\text{Product of roots = \dfrac{-(p^2+q^2)}{2} }$

Proof:

$:\longrightarrow\dfrac{1}{x+p}+\dfrac{1}{x+q}=\dfrac{1}{r}\\\\\text{So, taking LCM,}\\\\:\implies\dfrac{(x+q)+(x+p)}{(x+p)\times(x+q)}=\dfrac{1}{r}\\\\:\implies\dfrac{x+q+x+p}{x^2+px+qx+pq}=\dfrac{1}{r}\\\\:\implies\dfrac{2x+p+q}{x^2+(p+q)x+pq}=\dfrac{1}{r}\\\\\text{On cross multiplying}\\\\:\implies r(2x+p+q)=x^2+(p+q)x+pq\\\\:\implies2rx+pr+qr=x^2+(p+q)x+pq \\\\\text{Transposing LHS to RHS,}\\\\:\implies0=x^2+(p+q)x+pq-2rx-pr-qr\\\\:\implies x^2+(p+q-2r)x+(pq-pr-qr)=0$

$\text{We are given that, the sum of roots is 0}\\\\\text{This means \dfrac{-b}{a} =0, where b is coefficient of x and a is of x ^2 }\\\\\text{So,}\\\\:\implies \dfrac{-b}{a} = \dfrac{-(p+q-2r)}{1}\\\\:\implies0=-(p+q-2r)\\\\:\implies 0=p+q-2r\\\\:\implies p+q=2r\\\\:\implies r=\dfrac{p+q}{2}- - - -(1)\\\\\text{We know that, product of roots = \dfrac{c}{a} , c=constant}\\\\\text{So,}\\\\:\implies \dfrac{c}{a} =\dfrac{pq-pr-qr}{1}\\\\:\implies\text{Product of roots}=pq-pr-qr\\\\\text{From (1),}\\\\:\implies\text{Product of roots}=pq-p\left(\dfrac{p+q}{2}\right)-q\left(\dfrac{p+q}{2}\right)\\\\:\implies\text{Product of roots}=pq-\left(\dfrac{p^2+pq}{2}\right)-\left(\dfrac{pq+q^2}{2}\right)\\\\:\implies\text{Product of roots}=\dfrac{2pq-p^2-pq-pq-q^2}{2}\\\\:\implies\text{Product of roots}=\dfrac{2pq\!\!\!\!/\:\:-p^2-2pq\!\!\!\!/\:\:-q^2}{2}\\\\:\implies\text{Product of roots}=\dfrac{-p^2-q^2}{2} \\\\\bf{:\implies\text{\bf{Product of roots}}=\dfrac{-(p^2+q^2)}{2}}\\\\\text{\underline{HENCE PROVED}}$