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Mathematics 13 Online
OpenStudy (anonymous):

Solve for X. Check for extraneous answers. sqrt{x + sqrt{2x}}=sqrt{2}

OpenStudy (anonymous):

square both sides get \[x+\sqrt{2x}=2\] subtract \(x\) from both sides and square again \[\sqrt{2x}=2-x\] \[2x=(2-x)^2=4-4x+x^2\]

OpenStudy (anonymous):

then set equal to zero and solve the quadratic \[x^2-4x+4=2x\] \[x^2-6x+4=0\]

OpenStudy (anonymous):

would x= 2?

OpenStudy (anonymous):

oh no you have to either use the quadratic formula or complete the square

OpenStudy (anonymous):

and you also have to check your answer \[x^2-6x=-4\] \[(x-3)^2=-4+9=5\] \[x-3=\pm\sqrt5]\] \[x=3\pm\sqrt5\]

OpenStudy (anonymous):

now you have to check them it will turn out that \(3-\sqrt5\) works, while \(3+\sqrt5\) does not

OpenStudy (anonymous):

Thank you very much! Without you, I would have been stuck on this question for the whole night! Thank you!

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