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

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

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Answer 1 · 2015-07-21T09:51:56

Add $13$ $13$ to both sides to make the left hand side a perfect square trinomial, then take the square root of both sides and subtract $3$ $3$ to find:

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

Explanation:

Add $13$ $13$ to both sides to get:

$13 = x^{2} + 6 x + 9 = {(x + 3)}^{2}$ $13 = x^2+6x+9 = (x+3)^2$

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

Take the square root of both sides to get rid of the exponent:

$\to \sqrt{{(x + 3)}^{2}} = \pm \sqrt{13}$ $rarrsqrt((x+3)^2)=+-sqrt13$

So $x + 3 = \pm \sqrt{13}$ $x+3 = +-sqrt(13)$

Subtract $3$ $3$ from both sides to get:

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

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

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Explanation:

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

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