SOLUTION: &#8730;(x^2-x)+7=2(x^2-x)-1 The first part is all in the square root. I cannot get the answer at all. I keep comming up with no solution but my teacher says its not right.

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Question 598721: √(x^2-x)+7=2(x^2-x)-1
The first part is all in the square root.
I cannot get the answer at all. I keep comming up with no solution but my teacher says its not right.
Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website!
√(x^2-x)+7=2(x^2-x)-1
let u=(x^2-x)
√(u+7)=2u-1
square both sides
u+7=4u^2-4u+1
4u^2-5u-6=0
(4u+3)(u-2)=0
..
4u+3=0
u=-3/4
x^2-x=3/4
LCD:4
4x^2-4x=3
4x^2-4x-3=0
(2x-3)(2x+1)=0
x=3/2
or
x=-1/2
..
u-2=0
u=2
x^2-x=2
x^2-x-2=0
(x-2)(x+1)=0
x=2
or
x=-1
..
Note: 3/2 and -1/2 may be extraneous roots because we squared both sides. The original is a second degree equation which is only supposed to have 2 roots. You should check by plugging all four roots into the original equation and see which two works.

SOLUTION: &#8730;(x^2-x)+7=2(x^2-x)-1 The first part is all in the square root. I cannot get the answer at all. I keep comming up with no solution but my teacher says its not right.

SOLUTION: √(x^2-x)+7=2(x^2-x)-1 The first part is all in the square root. I cannot get the answer at all. I keep comming up with no solution but my teacher says its not right.