How do you solve using the completing the square method $x^{2} + 8 x - 9 = 0$ $x^2+8x-9=0$ ?

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Answer 1 · 2016-03-18T20:00:06

Completing the square is designed to give you a perfect square trinomial on the left of an equation.

Get any constants on the right side.
$x^{2} + 8 x = 9$ $x^2+8x=9$
Add ${(\frac{b}{2})}^{2} = {(\frac{8}{2})}^{2}$ $(b/2)^2=(8/2)^2$ to both sides
$x^{2} + 8 x + 16 = 9 + 16$ $x^2+8x+16=9+16$
Factor the left.
${(x + 4)}^{2} = 25$ $(x+4)^2=25$
Square root both sides.
$(x + 4) = \pm 5$ $(x+4)=+-5$
$x = {5 - 4, - 5 - 4} = {1, - 9}$ $x={5-4,-5-4} = {1,-9}$

How do you solve using the completing the square method x2+8x−9=0 x^2+8x-9=0?