Question

Find the roots of the equation by quadratic formula 1/2x-3 + 1/x-5 = 1​

Answer 1

$\huge{\mathfrak{\blue{\underline{\underline{Solution:-}}}}}$

$\sf{The\;given\;equation\;is,}$

$\sf{\frac{1}{2x-3}+\frac{1}{x-5}=1}$

$\sf{\implies \frac{(x-5)+(2x-3)}{(2x-3)\;(x-5)}=1}$

$\sf{\implies \frac{x-5+2x-3}{2x^{2}-10x-3x+15}=1}$

$\sf{\implies \frac{3x-8}{2x^{2}-13x+15}=1}$

$\sf{\implies 2x^{2}-13x+15=3x-8}$

$\sf{\implies 2x^{2}-13x-3x+15+8=0}$

$\sf{\implies 2x^{2}-16x+23=0}$

Comparing with standard equation $\sf{ax^{2}+bx+c=0,\;we\;get}$

$\sf{a=2\;;b=-16\;;c=23}$

$\sf{D=b^{2}-4ac=(-16)^{2}-4(2)(23)=256-184=72>0}$

So the given equation has real and distinct roots which are given by,    [QUADRATIC FORMULA].

$\sf{x=\frac{-b+\sqrt{D}}{2a}\;\;\;and\;\;\;x=\frac{-b-\sqrt{D}}{2a}}$

$\sf{x=\frac{16+\sqrt{72}}{4}\;\;\;and\;\;\; x=\frac{16-\sqrt{72}}{4}}$

$\sf{x=\frac{16+6\sqrt{2}}{4}\;\;\;and\;\;\;x=\frac{16-6\sqrt{2}}{4}}$

$\sf{x=\frac{8+3\sqrt{2}}{2}\;\;\;and\;\;\;x=\frac{8-3\sqrt{2}}{2}}$