Find the roots of the following quadratic equ | vedclass

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(D) To solve $x^{2}-8x+12=0$ by completing the square:
$1$. Move the constant term to the side: $x^{2}-8x = -12$
$2$. Add the square of half the coe

Vedclass · Accepted Answer

$2, 6$

Vedclass · Answer

(D) To solve $x^{2}-8x+12=0$ by completing the square:
$1$. Move the constant term to the side: $x^{2}-8x = -12$
$2$. Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $-8$, half of it is $-4$, and its square is $(-4)^{2} = 16$.
$3$. $x^{2}-8x+16 = -12+16$
$4$. Rewrite the left side as a perfect square: $(x-4)^{2} = 4$
$5$. Take the square root of both sides: $x-4 = \pm 2$
$6$. Solve for $x$: $x = 4+2 = 6$ or $x = 4-2 = 2$.
Thus, the roots are $2$ and $6$.

Find the roots of the following quadratic equation by the method of completing the square: $x^{2}-8x+12=0$

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