How do you solve using the completing the square method $x^{2} + 8 x = 7$ $x^2 +8x = 7$ ?

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Answer 1 · 2016-03-20T00:16:19

Add $16$ $16$ to both sides to "complete the square" on the left hand side and hence find:

$x = - 4 \pm \sqrt{23}$ $x = -4+-sqrt(23)$

One way is as follows:

First add $16$ $16$ to both sides to get:

$x^{2} + 8 x + 16 = 23$ $x^2+8x+16 = 23$

Note that $x^{2} + 8 x + 16 = {(x + 4)}^{2}$ $x^2+8x+16 = (x+4)^2$ , so we have:

${(x + 4)}^{2} = 23$ $(x+4)^2 = 23$

Therefore:

$x + 4 = \pm \sqrt{23}$ $x+4 = +-sqrt(23)$

Hence:

$x = - 4 \pm \sqrt{23}$ $x = -4+-sqrt(23)$

How do you solve using the completing the square method x2+8x=7x^2 +8x = 7?