$$ \sqrt{a-\sqrt{a+x}}=x $$ This equation contains one variable x we have to find the value of x.I tried to simplify it but it doesn't work....i have also tried the basic concepts of quadratics but it is just waste
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Is it $\sqrt{a-(\sqrt{a}+x)}=x$ or $\sqrt{a-\sqrt{a+x}}=x$? – Spenser Jun 21 '16 at 13:33
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Is it correct my edit? – Emilio Novati Jun 21 '16 at 13:35
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@EmilioNovati The way it was written is much more likely to be the first option of my comment. – Spenser Jun 21 '16 at 13:39
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Square both sides to get rid of the square root: $$a-\sqrt{a+x}=x^2$$ Isolate the radical: $$\sqrt{a+x}=a-x^2$$ Square both sides to get rid of the square root: $$a+x=x^4-2ax^2+a^2$$ Set the polynomial equal to $0$: $$0=x^4-2ax^2-x+(a^2-a)$$ Now, this is a depressed quartic, which can be solved using Ferrari's Solution or any other method of solving depressed quartics. Once you find the four solutions to this polynomial, plug them back into the original equation to make sure there are no extraneous solutions.
Noble Mushtak
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If the question is $\sqrt{a-(\sqrt{a}+x)}=x$.
Then let $t=\sqrt{a}+x $. So $ a-t=x^2$ i.e $ t=(\sqrt{a}-x)(\sqrt{a}+x)$. i.e $1=\sqrt{a}-x$
Therefore $$x=\sqrt{a}-1$$
A s
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