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

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Answer 1 · 2015-05-23T13:07:58

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

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

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

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

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

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

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

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

$x-4 = +-sqrt(9) = +-3$

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

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