How do you solve by completing the square for $x^2-8x+7$ ?

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Answer 1 · 2015-07-07T21:35:21

You have to take half the $x$ -coefficient (=-4) and square it. At the same time you have to balance the 7 that is allready there.

$x^2-8x+16=(x-4)^2$

So we need to add $9$ to both sides:
$x^2-8x+16=9->$
$(x-4)^2=9->$
So: $x-4=sqrt9=3->x=7$
Or: $x-4=-sqrt9=-3->x=1$

How do you solve by completing the square for x^2-8x+7x^2-8x+7?