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How do I solve a problem with 3 radicals.
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\[\sqrt{x+\sqrt{2x}} = \sqrt{2x}\]
start by squaring both sides
\[(\sqrt{x+\sqrt{2x}})^{2} = (\sqrt{2x})^{2}\] like this?
yes cancel the radical
\[x+\sqrt{2x} = 2x\] Now what do I do?
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subtract x both sides
Would it be\[\sqrt{2x} = 2x-x\]
To give me \[\sqrt{2x} = x\]
yes ! square it again
Then I would have,\[2x = x^{2}\] is this true?
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Correct !
its a quadratic, u can factor it and solve x
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