How do you solve $x^{2} - 6 x - 4 = 0$ $x^2-6x-4=0$ by completing the square?

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How do you solve $x^{2} - 6 x - 4 = 0$ $x^2-6x-4=0$ by completing the square?

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Answer 1 · 2015-07-16T18:06:53

$x = 3 + \sqrt{13}, 3 - \sqrt{13}$ $x=3+sqrt13, 3-sqrt13$

Explanation:

$x^{x} - 6 x - 4 = 0$ $x^x-6x-4=0$

Completing the square means that we will force a perfect square trinomial on the left side of the equation, then solve for $x$ $x$ .

Add $4$ $4$ to both sides of the equation.

$x^{2} - 6 x = 4$ $x^2-6x=4$

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

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

$x^{2} - 6 x + 9 = 4 + 9$ $x^2-6x+9=4+9$ =

$x^{2} - 6 x + 9 = 13$ $x^2-6x+9=13$

We now have a perfect square trinomial in the form $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ $a^2-2ab+b^2=(a-b)^2$ , where $a = x and b = 3$ $a=x and b=3$ .

Substitute ${(x - 3)}^{2}$ $(x-3)^2$ for $x^{2} - 6 x + 9$ $x^2-6x+9$ .

${(x - 3)}^{2} = 13$ $(x-3)^2=13$

Take the square root of both sides.

$x - 3 = \pm \sqrt{13}$ $x-3=+-sqrt13$

Add $3$ $3$ to both sides of the equation.

$x = 3 \pm \sqrt{13}$ $x=3+-sqrt13$

Solve for $x$ $x$ .

$x = 3 + \sqrt{13}$ $x=3+sqrt13$

$x = 3 - \sqrt{13}$ $x=3-sqrt13$

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How do you solve $x^{2} - 6 x - 4 = 0$ $x^2-6x-4=0$ by completing the square?

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Explanation:

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How do you solve x^2-6x-4=0x2−6x−4=0x^2-6x-4=0 by completing the square?

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How do you solve $x^{2} - 6 x - 4 = 0$ $x^2-6x-4=0$ by completing the square?