How do you complete the square to solve $x^{2} + 2 x - 10 = 14$ $x^2+2x-10=14$ ?

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Answer 1 · 2015-05-23T20:26:45

If you first add $11$ $11$ to both sides you get:

$x^{2} + 2 x + 1 = 25$ $x^2+2x+1 = 25$

Now

$x^{2} + 2 x + 1 = (x + 1) (x + 1) = {(x + 1)}^{2}$ $x^2+2x+1 = (x+1)(x+1) = (x+1)^2$

and $25 = 5^{2}$ $25 = 5^2$

So we have

${(x + 1)}^{2} = 5^{2}$ $(x+1)^2 = 5^2$

Hence

$x + 1 = \pm \sqrt{5^{2}} = \pm 5$ $x+1 = +-sqrt(5^2) = +-5$

Subtracting 1 from both sides we get:

$x = - 1 \pm 5$ $x = -1 +-5$

That is $x = 4$ $x = 4$ or $x = - 6$ $x=-6$

How do you complete the square to solve x2+2x−10=14x^2+2x-10=14?