Question

find the roots of the given quadratic equation by completing the square method
x square - 6 X + 8 = 0

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Answer 1

Given :

• Quadratic equation : x² - 6x + 8 = 0

To Find :

• Roots of the given quadratic equation

Solution :

➨ x² - 6x + 8 = 0

➨ x² - 6x = - 8

By adding 3² on both side

➨ x² - 6x + 3² = - 8 + 3²

➨ x² - 6x + 3² = - 8 + 9

Using identity : ( a - b )² = a² - 2ab + b²

➨ ( x - 3 )² = 1

➨ x - 3 = √1

➨ x - 3 = ± 1

➨ when x - 3 = 1 , x = 4

➨ when x - 3 = - 1 , x = 2

Answer 2

$\huge{\underline{\underline{\pink{Ans}\red{wer:}}}}$

$\bold{Roots \ are \ 4 \ and \ 2.}$

$\huge\orange{\underline{\underline{Given:}}}$

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

$\bold{=>x^{2}-6x+8=0}$

$\bold\purple{\underline{\underline{To \ find:}}}$

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

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

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

$\bold{=>x^{2}-6x+8=0}$

$\bold{Here,}$

$\bold{Coefficient \ of \ x^{2}=1}$

$\bold{Coefficient \ of \ x=-6}$

$\bold{Constant \ term=8}$

$\bold{=>x^{2}-6x=-8}$

______________________________

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

$\bold{=>[\frac{1}{2}×(-6)]^{2}}$

$\bold{=>-3^{2}}$

$\bold{=9}$

$\bold{Add \ 9 \ on \ both \ sides \ of \ equation}$

$\bold{=>x^{2}-6x+9=-8+9}$

$\bold{=>(x-3)^{2}=1}$

$\bold{On \ taking \ square \ root \ of \ both \ sides}$

$\bold{=>x-3=1 \ or \ -1}$

$\bold{\therefore{x=1+3 \ or \ -1+3}}$

$\bold{\therefore{x=4 \ or \ 2.}}$

$\bold\purple{\tt{\therefore{Roots \ are \ 4 \ and \ 2.}}}$