How do you complete the square to solve $x^{2} - 2 x = 15$ $x^2 - 2x = 15$ ?

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Answer 1 · 2015-06-02T23:20:08

Given $x^{2} - 2 x = 15$ $x^2-2x=15$ add 1 to both sides to get:

$16 = x^{2} - 2 x + 1 = {(x - 1)}^{2}$ $16 = x^2-2x+1 = (x-1)^2$

So $x - 1 = \pm \sqrt{16} = \pm 4$ $x-1 = +-sqrt(16) = +-4$

Add $1$ $1$ to both sides to get:

$x = 1 \pm 4$ $x = 1+-4$

That is: $x = - 3$ $x = -3$ or $x = 5$ $x=5$

How do you complete the square to solve x2−2x=15x^2 - 2x = 15?