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 · 2015-07-05T05:39:09

Force a perfect square trinomial on the left side of the equation. Take the square root of both sides and solve for $x$ $x$ .

Explanation:

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

Force a perfect square trinomial such that $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ $a^2+2ab+b^2=(a+b)^2$ on the left side of the equation. Then take the square root of both sides and solve for $x$ $x$ .

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

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

Take the coefficient of the $x$ $x$ term and divide it by $2$ $2$ and square the result.

$\frac{6}{2} = 3;$ $6/2=3;$ $3^{2} = 9$ $3^2=9$

Add $9$ $9$ to both sides.

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

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

$a = x;$ $a=x;$ $b = 3$ $b=3$

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

Take the square root of both sides.

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

Subtract $3$ $3$ from both sides.

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

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

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

1 Answer

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

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Explanation:

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