Solve: ___ ______ √x-2 -1 = √2(x-3)

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Solve: ___ ______ √x-2 -1 = √2(x-3)

Mathematics
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square both sides to get (x-2)-2\[\sqrt{x-2}\]+1=2x-6 x-1-2\[\sqrt{x-2}/]=2x-6 2\[\sqrt{x-2}\]=-x+5 \[\sqrt{x-2}\]=-x+5/2 square both sides again to get x-2=(-x+5/2)^2 x=(-x+5/2)^2+2 x=(x^2-10x+25)/4 +2 4x=x^2-10x+27 x^2-14x+27=0 quadratic from there
\[\sqrt{x-2} -1 = \sqrt{2(x-3)}\] \[(\sqrt{x-2})^{2} - (1)^{2} = \sqrt{2x-6}\] \[x-2-1 = 2x-6 \] \[x-3 = 2x-6\] \[-x = -3\] x=3
no you have o square te entire leftside so its (sqrt(x-2)-1)^2 so you get -2sqrt(x-2)+(x-2)+1=-2sqrt(x-2)+x-1

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Other answers:

hmm ok, my instructors answer is 3...trying to figure it out by your method
problem is that you cant square the terms individualy to clear the sqrt
Hmm.. it is x^2-14x-33=0
doing itnot in my head at 2 am i got x^2-8x+12=0 but still not geting to "3"
i appreciate the help
holy hell dropped the 2 on the right side gimme one sec
@denebel has the correct quadratic and it also has one answer that is 3
x+5-2*sqrt (x-2) = 2x x+5 = 2x + 2*sqrt (x-2) 5 = 2x + 2*sqrt (x-2) - x 5-x = 2*sqrt (x-2) (5-x)^2 = (2*sqrt (x-2))^2 x^2 - 10x + 25 = 4x - 8 0 = -x^2 + 14x + 33 0 = x^2 - 14x - 33
x=3, x=11. Then you check each value by plugging it back into the original equation. Only x=3 checks out, therefore, x=3 is your answer.
Oops Typo 0 = -x^2 + 14x - 33 0 = x^2 - 14x + 33
your awesome, thank you... now to break it down so i learn it lol

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