Question 60107: How do you complete the square of x2-8x=33 and x2-8x+40=0 Found 2 solutions by checkley71, ankor@dixie-net.com:Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website! X^2-8X=33
X^2-8X+16=33+16
(X-4)(X-4)=49
X-4=7
X=7+4
X=11
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X^2-8X+40=0
X^2-8X=-40
X^2-8X+16=-40+16
(X-4)(X-4)=-24
X-4=2SQRT6
You can put this solution on YOUR website! Complete the square by filling in the blank to make it a perfect square:
x^2 - 8x + ___ = 33
:
Take the coefficient of x, divide it by 2 and square it:
Coefficient is -8, divided by 2 is -4, squaring 4 gives you + 16
Remember if you add 16 to the left side of the equation you must do the same to the right side.
x^2 - 8x + 16 = 33 + 16
:
x^2 - 8x + 16 = 49
:
Since we made it a perfect square we can write it:
(x - 4)^2 = 49
:
Find the square root of both sides:
(x - 4) = +/-7
:
x = +4 + 7
x = +11
and
x = +4 - 7
x = -3
:
The solution is x = +11 and x = -3
:
:
x^2 - 8x + 40 = 0
:
Subtract 40 from both sides and complete the square:
x^2 - 8x + ___ = -40
:
Since the coefficient of x is 8 again. we will have:
x^2 - 8x + 16 = -40 + 16
:
x^2 - 8x + 16 = -24
:
(x -4)^2 = - 24
:
x - 4 = +/-SqRt(-24)
:
Using the i factor (which is the square root of -1)
x - 4 = +/- i*SqRt(+24)
:
x - 4 = +/- i*SqRt(4*6)
:
Extract 2 by taking out the square root of 4
x - 4 = +/- 2i*SqRt(6)
:
x = 4 + 2i*SqRt(6)
and
x = 4 - 2i*SqRt(6)
:
Did this make sense to you??
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