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#