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How do you solve using the completing the square method $x^{2} - 8 x + 7 = 0$ $x^2-8x+7=0$ ?

1 Answer

Sep 20, 2016

$x = 7 XX or XX x = 1$ $x=7color(white)("XX")orcolor(white)("XX")x=1$

Explanation:

Given
$XXX x^{2} - 8 x + 7 = 0 XX$ $color(white)("XXX")x^2-8x+7=0color(white)("XX")$ Note: I modified the question to include the $= 0$ $color(blue)(=0)$

Then
$XXX x^{2} - 8 x = - 7$ $color(white)("XXX")x^2-8x=-7$

Remembering the general relation:
$XXX x^{2} + 2 a x + a^{2} = {(x + a)}^{2}$ $color(white)("XXX")x^2+2ax+a^2 = (x+a)^2$

If $x^{2} - 8 x$ $x^2-8x$ are to be the first two terms of $x^{2} + 2 a x + a^{2}$ $x^2+2ax+a^2$
then
$XXX a = - 4$ $color(white)("XXX")a=-4$
and
$XXX a^{2} = 16$ $color(white)("XXX")a^2=16$

So we will need to add $16$ $16$ (to both sides) to $underline("complete the square")$

$color(white)("XXX")x^2-8x+16 = -7+16$

$color(white)("XXX")(x-4)^2=9$

$color(white)("XXX")x-4=+-3$

$color(white)("XXX")x=4+-3$

$color(white)("XXX")x=7color(white)("XX")orcolor(white)("XX")x=1$

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