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

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Answer 1 · 2015-07-07T02:18:53

Solve y = x^2 - 8x - 9 = 0

In this case, you don't need to solve by completing the squares.
Use the shortcut.
When (a - b + c = 0) --> 2 real roots: (-1) and (-c/a = 9).

Answer 2 · 2015-07-07T21:27:54

You take half the $x$ -coefficient ( $-8->-4$ ) and square it ( $16$ )

This means you have to add $16$ to both sides of the equation:
$x^2-8x+16=9+16->$
$(x-4)^2=25->$
So: $x-4=sqrt25=5->x=9$
Or: $x-4=-sqrt25=-5->x=-1$

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