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

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

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Answer 1 · 2015-07-19T21:32:35

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

Explanation:

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

Completing the square means forcing a perfect square trinomial on the left side of the equation.

Bring all terms to the left side.

$- x^{2} + 6 x + 4 = 0$ $-x^2+6x+4=0$

Multiply boths sides times $- 1$ $-1$ .

$- 1 (- x^{2} + 6 x + 4 = 0)$ $-1(-x^2+6x+4=0)$ =

$x^{2} - 6 x - 4 = 0$ $x^2-6x-4=0$

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

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

Divide the x-term by 2 and square the result. Add 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 on the left side. $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ $a^2-2ab+b^2=(a-b)^2$ , where $a = x and b = 3$ $a=x and b=3$ .

Substitute ${(x - 3)}^{2}$ $(x-3)^2$ into the equation.

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

Take the square root of both sides and solve for $x$ $x$ .

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

Add $3$ $3$ to both sides.

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

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 $6 x + 4 = x^{2}$ $6x + 4 = x^2$ by completing the square?

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