How do you solve $x^2+6x=7$ using completing the square?

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Answer 1 · 2015-06-19T22:14:37

$(x+3)^2 = x^2+6x+9$

So: $(x+3)^2 = 7 + 9 = 16 = 4^2$

Hence $x = -7$ or $x = 1$

$(x+3)^2 = x^2+6x+9$

So: $(x+3)^2 = x^2+6x+9 = 7 + 9 = 16 = 4^2$

Take square root of both ends, allowing for both possibilities:

$x+3 = +-sqrt(4^2) = +-4$

Subtract $3$ from both sides to get:

$x = -3+-4$

That is $x = -7$ or $x = 1$

How do you solve x^2+6x=7 x^2+6x=7 using completing the square?