How do you solve the quadratic equation by completing the square: $x^{2} - 6 x - 8 = 0$ $x^2-6x-8=0$ ?

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How do you solve the quadratic equation by completing the square: $x^{2} - 6 x - 8 = 0$ $x^2-6x-8=0$ ?

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Answer 1 · 2015-07-14T23:00:50

$x = 3 + \sqrt{17}, 3 - \sqrt{17}$ $x=3+sqrt(17), 3-sqrt(17)$

Explanation:

$x^{2} - 6 x - 8 = 0$ $x^2-6x-8=0$

In order to solve a quadratic equation by completing the square, we must force a perfect square trinomial on the left side. $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ $a^2+2ab+b^2=(a+b)^2$

Add $8$ $8$ to both sides of the equation.

$x^{2} - 6 x = 8$ $x^2-6x=8$

Divide the coefficient of the $x$ $x$ term by $2$ $2$ and square the result. Add it to both sides of the equation.

${(\frac{- 6}{2})}^{2} = {(- 3)}^{2} = 9$ $((-6)/(2))^2=(-3)^2=9$

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

$x^{2} - 6 x + 9 = 17$ $x^2-6x+9=17$

We now have a perfect square trinomial on the left side, in which
$a = x,$ $a=x,$ and $b = - 3$ $b=-3$ .

Now we can solve for $x$ $x$ .

${(x - 3)}^{2} = 17$ $(x-3)^2=17$

Take the square root of both sides.

$(x - 3) = \pm \sqrt{17}$ $(x-3)=+-sqrt(17)$

$x = 3 \pm \sqrt{17}$ $x=3+-sqrt(17)$

$x = 3 + \sqrt{17}$ $x=3+sqrt(17)$

$x = 3 - \sqrt{17}$ $x=3-sqrt(17)$

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How do you solve the quadratic equation by completing the square: $x^{2} - 6 x - 8 = 0$ $x^2-6x-8=0$ ?

1 Answer

Explanation:

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Science

Math

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How do you solve the quadratic equation by completing the square: x2−6x−8=0x^2-6x-8=0?

1 Answer

Explanation:

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How do you solve the quadratic equation by completing the square: $x^{2} - 6 x - 8 = 0$ $x^2-6x-8=0$ ?