Find the roots of the following quadratic equ | vedclass

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(B) Given equation: $x^{2}-4x-8=0$Step $1$: Shift the constant term to the right side: $x^{2}-4x=8$Step $2$: Add the square of half the coefficient of

Vedclass · Accepted Answer

$2+2\sqrt{3}, 2-2\sqrt{3}$

Vedclass · Answer

(B) Given equation: $x^{2}-4x-8=0$Step $1$: Shift the constant term to the right side: $x^{2}-4x=8$Step $2$: Add the square of half the coefficient of $x$ to both sides. The coefficient of $x$ is $-4$,so half of it is $-2$,and its square is $(-2)^{2}=4$.$x^{2}-4x+4=8+4$Step $3$: Write the left side as a perfect square: $(x-2)^{2}=12$Step $4$: Take the square root on both sides: $x-2 = \pm\sqrt{12}$$x-2 = \pm 2\sqrt{3}$Step $5$: Solve for $x$: $x = 2 \pm 2\sqrt{3}$Thus,the roots are $2+2\sqrt{3}$ and $2-2\sqrt{3}$.

Find the roots of the following quadratic equation by the method of completing the square: $x^{2}-4x-8=0$

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