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Solve for X. Check for extraneous answers. sqrt{x + sqrt{2x}}=sqrt{2}
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square both sides get \[x+\sqrt{2x}=2\] subtract \(x\) from both sides and square again \[\sqrt{2x}=2-x\] \[2x=(2-x)^2=4-4x+x^2\]
then set equal to zero and solve the quadratic \[x^2-4x+4=2x\] \[x^2-6x+4=0\]
would x= 2?
oh no you have to either use the quadratic formula or complete the square
and you also have to check your answer \[x^2-6x=-4\] \[(x-3)^2=-4+9=5\] \[x-3=\pm\sqrt5]\] \[x=3\pm\sqrt5\]
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now you have to check them it will turn out that \(3-\sqrt5\) works, while \(3+\sqrt5\) does not
Thank you very much! Without you, I would have been stuck on this question for the whole night! Thank you!
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