Question

By the method of completing squares, find the roots of the quadratic equation : 2x2 - 2root2x + 1 = 0

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• 2x²-2√2x+1

${\sf{\green{\underline{\large{To\:find}}}}}$

• By completing squares method

${\sf{\pink{\underline{\Large{Explanation}}}}}$

☞2x²-2√2x+1

Multiplying By 2

=>2(2x²-2√2x+1)

=>4x²-4√2+2

=>(2x)²-2×2x × √2+(√2)²-(√2)²+1

=>(2x-√2)²-2+1

=>(2x-√2)²=1

=>(2x-√2) =√1

Opening sign with +

2x-√2=1

=>2x=1+√2

=>X=(1+√2)/2

Opening sign with -

2x-√2=-1

=>2x=-1+√2

=>X=(-1+√2)/2

Answer 2

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

$\bold{Roots \ are \frac{-1}{\sqrt2} \ and \frac{-1}{\sqrt2}}$

$\bold\orange{Given:}$

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

$\bold{2x^{2}-2\sqrt2x+1=0}$

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

$\bold{Roots \ of \ the \ quadratic \ equation}$

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

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

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

$\bold{2x^{2}-2\sqrt2x+1=0}$

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

$\bold{=>x^{2}-\sqrt2x+\frac{1}{2}=0}$

$\bold{=>x^{2}-\sqrt2x=\frac{-1}{2}}$

___________________________________

$\bold{=>[\frac{1}{2}×Coefficient \ of \ x]^{2}}$

$\bold{=>[\frac{1}{2}×-\sqrt2]^{2}}$

$\bold{=>(\frac{-\sqrt2}{2})^{2}}$

$\bold{=>\frac{2}{4}}$

$\bold{=>\frac{1}{2}}$

___________________________________

$\bold{Adding \frac{1}{2} \ on \ both \ sides,}$

$\bold{we \ get}$

$\bold{=>x^{2}-\sqrt2x+\frac{1}{2}=\frac{-1}{2}+\frac{1}{2}}$

$\bold{=>(x+\frac{1}{\sqrt2})^{2}=0}$

$\bold{we \ get}$

$\bold{=>(x+\frac{1}{\sqrt2})(x+\frac{1}{\sqrt2})=0}$

$\bold{=>x=\frac{-1}{\sqrt2} \ or \frac{-1}{\sqrt2}}$

$\bold\purple{\tt{\therefore{Root \ are \ \frac{-1}{\sqrt2} \ and \frac{-1}{\sqrt2}.}}}$