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

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Answer 1 · 2015-03-31T00:40:25

The quadratic that is a square and starts with the terms $x^{2}$ $x^2$ and $8 x$ $8x$
is $x^{2} + 8 x + 16$ $x^2+8x+16$

Re-write the given equation with the square quadratic at the beginning
$x^{2} + 8 x + 16 - 25 = 0$ $x^2+8x+16-25 = 0$

${(x + 4)}^{2} = 25$ $(x+4)^2 = 25$

$x + 4 = \pm \sqrt{25}$ $x+4 = +-sqrt(25)$

$x = - 4 \pm 5$ $x=-4+-5$

$x = - 9$ $x=-9$ or $x = 1$ $x=1$

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