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

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

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Answer 1 · 2015-05-23T13:07:58

Given $y = x^{2} - 8 x + 7$ $y=x^2-8x+7$

Note that if $x^{2} - 8 x$ $x^2-8x$ are the first two term of a squared expression ${(x + a)}^{2}$ $(x+a)^2$
then $a = \frac{8}{2} = 4$ $a= 8/2 = 4$ and $a^{2} = 16$ $a^2 = 16$

$y = x^{2} - 8 x + 7$ $y= x^2 -8x+7$

$= x^{2} - 8 x + 4^{2} + 7 - 16$ $= x^2-8x + color(red)(4^2) + 7 color(red)(-16)$

$= {(x - 4)}^{2} - 9$ $=(x-4)^2 -9$

Now, the real problem
What do you mean by solve and equation of the form $y =$ $y =$ and expression in terms of $x$ $x$ ?

I will assume you wish to determine the value(s) of $x$ $x$ for which $y = 0$ $y=0$ (although this is not obvious).

If ${(x - 4)}^{2} - 9 = 0 (= y)$ $(x-4)^2-9 = 0 (=y)$
then
${(x - 4)}^{2} = 9$ $(x-4)^2 = 9$

$x - 4 = \pm \sqrt{9} = \pm 3$ $x-4 = +-sqrt(9) = +-3$

and
$x = 7$ $x= 7$ or $x = 1$ $x=1$

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

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