How do you solve $x^{2} = 8 x - 15$ $x^2= 8x - 15$ by completing the square?

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Answer 1 · 2017-07-11T01:46:32

Rearranging first

$x^{2} - 8 x + 15 = 0$ $x^2-8x+15=0$

${(x - 4)}^{2} - 4^{2} + 15 = 0$ $(x-4)^2-4^2+15=0$

${(x - 4)}^{2} - 1 = 0$ $(x-4)^2-1=0$

${(x - 4)}^{2} = 1$ $(x-4)^2=1$

$(x - 4) = \pm 1$ $(x-4) = +-1$

$x = 4 \pm 1$ $x=4+-1$

This gives two values of x, one where you add the 1 and one where you subtract 1

Answer 2 · 2017-07-12T09:31:41

$x = 3 or x = 5$ $x =3 or x =5$

$x^{2} = 8 x - 15$ $x^2 = 8x -15$

$x^{2} - 8 x = - 15 \leftarrow$ $x^2 -8x = -15" "larr$ leave the constant on the right side

Make a perfect square by adding ${(\frac{b}{2})}^{2}$ $(b/2)^2$ to both sides
$b = - 8 \to {(\frac{- 8}{2})}^{2} = 16$ $b =-8" "rarr ((-8)/2)^2 =16$

$x^{2} - 8 x + 16 = - 15 + 16$ $x^2-8x +16 = -15+16$

${(x - 4)}^{2} = 1$ $(x-4)^2 = 1$

$x - 4 = \pm 1$ $x-4 = +-1$

$x - 4 = 1 \to x = 5$ $x -4 = 1" "rarr x =5$

$x - 4 = - 1 \to x = 3$ $x-4 = -1" "rarr x = 3$

How do you solve x2=8x−15x^2= 8x - 15 by completing the square?