How do you solve by completing the square: $x^{2} - 8 x - 1 = 0$ $x^2 - 8x - 1 = 0$ ?

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How do you solve by completing the square: $x^{2} - 8 x - 1 = 0$ $x^2 - 8x - 1 = 0$ ?

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Answer 1 · 2015-03-30T01:58:13

If $x^{2} - 8 x$ $x^2-8x$ are the first 2 terms of a squared expression of the form ${(x + a)}^{2}$ $(x+a)^2$
then the third term should be
${(\frac{- 8}{2})}^{2} = 16$ $((-8)/2)^2 = 16$

If we are going to replace
$x^{2} - 8 x - 1$ $x^2-8x-1$
with
$x^{2} - 8 x + 16$ $x^2-8x+16$
we will need to add $+ 17$ $+17$ to both sides of the equation

$x^{2} - 8 x + 16 = 17$ $x^2-8x+16 = 17$

re-write the left-side as a square
${(x - 4)}^{2} = 17$ $(x-4)^2 = 17$

then take the square root of both sides
$x - 4 = \pm \sqrt{17}$ $x-4 = +-sqrt(17)$

$\to x = 4 + \sqrt{17}$ $rarr x = 4+sqrt(17)$ or $x = 4 - \sqrt{17}$ $x = 4 -sqrt(17)$

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How do you solve by completing the square: $x^{2} - 8 x - 1 = 0$ $x^2 - 8x - 1 = 0$ ?

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