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Question 104663: x-2*sqrt(x-3)=3
solve for zero...
I subtract x from both sides which =
-2*sqrt(x-3)=3-x
then I divide by -2 from both sides
sqrt(x-3)=(3-x)/-2
then to get rid of the "sqrt" ^2 on both sides
x-3=(9+x^2)/4
then I subtract "x" and add 3 to both sides to solve for zero.
0=(9+x^2)/4-x+3
then I find a common denominator of 4 and solve
0=(9+x^2)/4 -4x/4+12/4 which equals 0=(x^2-4x+21)/4
then I multiply 4/1 on both side the loose the denominator.
0=x^2-4x+21 which can not be factored, but my text book gives a solution set of {7,3}. I cant find were I went wrong, please help
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! x-2*sqrt(x-3)=3
solve for zero...
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x-2*sqrt(x-3)=3
solve for zero...
-2sqrt(x-3) = 3-x
Square both sides to get:
4(x-3) = 9-6x+x^2
4x-12 = 9-6x+x^2
x^2-10x+21=0
(x-3)(x-7)=0
x = 3 or x=7
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Cheers,
Stan H.
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