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

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

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Answer 1 · 2015-07-07T21:35:21

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

Explanation:

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

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

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

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