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How do you solve using the completing the square method $x^{2} + 8 x = 9$ $x^2+8x=9$ ?

1 Answer

Feb 28, 2016

$x = 1$ $x=1$ or $x = 9$ $x=9$
(see below for method of completing the square).

Explanation:

A general squared binomial has the relation:
$XXX {(a + b)}^{2} = a^{2} + 2 a b + b^{2}$ $color(white)("XXX")(a+b)^2=a^2+2ab+b^2$

If $x^{2} + 8 x$ $x^2+8x$ are the first two terms of an expanded squared binomial
then $a = 1$ $a=1$ and $b = 4$ $b=4$
and the third term needed to complete the square would be $b^{2} = 16$ $b^2=16$ .

If we add $16$ $16$ to the left side to complete the square there
we will need to add $16$ $16$ to the right side to maintain the equality.
$XXX x^{2} + 8 x + 16 = 9 + 16$ $color(white)("XXX")x^2+8x+16=9+16$

Re-writing the left side as a squared binomial and simplifying the right side:
$XXX {(x + 4)}^{2} = 25$ $color(white)("XXX")(x+4)^2=25$

If we take the square root of both sides:
$XXX (x + 4) = \pm 5$ $color(white)("XXX")(x+4)=+-5$

which implies:
either $x = 1$ $x=1$ or $x = - 9$ $x=-9$

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