Question

2x^2+5x-6=0solve by completing square method

Answer 1

$\bold\red{\underline{\underline{Answer:}}}$

$\bold{Roots \ are \frac{-5+\sqrt73}{4} \ and \frac{-5-\sqrt73}{4}}$

$\bold\orange{Given:}$

$\bold{The \ given \ quadratic \ equation \ is}$

$\bold{=>2x^{2}+5x-6=0}$

$\bold\pink{To \ find:}$

$\bold{=>Roots \ of \ the \ equation \ by \ completing}$

$\bold{the \ square \ method.}$

$\bold\green{\underline{\underline{Solution:}}}$

$\bold{By \ completing \ the \ square \ method}$

$\bold{=>2x^{2}+5x-6=0}$

$\bold{Dividing \ the \ equation \ by \ 2, \ we \ get,}$

$\bold{=>x^{2}+\frac{5x}{2}-3=0}$

$\bold{=>x^{2}+\frac{5x}{2}=3}$

_____________________________

$\bold{(\frac{1}{2}×Coefficient \ of \ x)^{2}}$

$\bold{=>(\frac{1}{2}×\frac{5}{2})^{2}}$

$\bold{=>(\frac{5}{4})^{2}}$

$\bold{=>\frac{25}{16}}$

_____________________________

$\bold{Adding \frac{25}{16} \ on \ both \ sides}$

$\bold{=>x^{2}+\frac{5x}{2}+\frac{25}{16}=3+\frac{25}{16}}$

$\bold{=>(x+\frac{5}{4})^{2}=\frac{48+25}{16}}$

$\bold{=>(x+\frac{5}{4})^{2}=\frac{73}{16}}$

$\bold{Taking \ square \ root \ of \ both \ sides}$

$\bold{=>x+\frac{5}{4}=\frac{\sqrt73}{4} \ or \frac{-\sqrt73}{4}}$

$\bold{=>x=\frac{\sqrt73}{4}-\frac{5}{4} \ or \frac{-\sqrt73}{4}-\frac{5}{4}}$

$\bold{x= \frac{-5+\sqrt73}{4}}$

$\bold{or \ x=\frac{-5-\sqrt73}{4}}$

$\bold\purple{\tt{\therefore{Roots \ are \frac{-5+\sqrt73}{4} \ and \frac{-5-\sqrt73}{4}}}}$