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Mathematics 19 Online
OpenStudy (anonymous):

Solve. Check for extraneous solutions. sqrt(x+srt2x)=sqrt2x. Please show step by step instruction. Thankyou.

OpenStudy (anonymous):

$$\sqrt{x+\sqrt{2x}}=\sqrt{2x}$$is that correct?

OpenStudy (anonymous):

yeah.

OpenStudy (anonymous):

@oldrin.bataku

OpenStudy (anonymous):

okay, first thing we want to do is square both sides to get rid of the first radicals:$$\sqrt{x+\sqrt{2x}}=\sqrt{2x}\\\left(\sqrt{x+\sqrt{2x}}\right)^2=\left(\sqrt{2x}\right)^2\\x+\sqrt{2x}=2x\\\sqrt{2x}=2x-x\\\sqrt{2x}=x$$now we square again:$$\left(\sqrt{2x}\right)^2=x^2\\2x=x^2\\x^2-2x=0\\x(x-2)=0$$

OpenStudy (anonymous):

to verify these solutions are not extraneous, plug them back in.

OpenStudy (anonymous):

How? And thank you

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