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

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Answer 1 · 2017-07-11T01:46:32

Rearranging first

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

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

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

$(x-4)^2=1$

$(x-4) = +-1$

$x=4+-1$

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

$x =3 or x =5$

$x^2 = 8x -15$

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

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

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

$(x-4)^2 = 1$

$x-4 = +-1$

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

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

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