Please, help me to solve this equation:


I think if we continue to square, it would have to face a high-order equation.

I guess $x = 30$, but do not know how to solve.

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    $\begingroup$ One can not but notice in L.H.S. the number under the all square-roots is 5 and it is the number on the R.H.S. , repeatedly substituting we get an infinite run of square-roots over x , √(x-√(x-√..) = 5 , squaring both sides and noticing that only one x gets out we obtain x - 5 = 5^2 = 25 $\endgroup$ – Souvik Dey Oct 18 '12 at 7:22
  • $\begingroup$ @SouvikDey Sorry, I don't understand everything in your comment. Could you give more details? $\endgroup$ – vesszabo Oct 18 '12 at 8:30
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    $\begingroup$ The logic is similar to $\sqrt{x-5} =5 \implies \sqrt{x-\sqrt{x-5}} = 5$ $\endgroup$ – user14082 Oct 18 '12 at 8:35
  • $\begingroup$ @JayeshBadwaik But it seems to me that Souvik Dey uses it in the opposite direction. $\endgroup$ – vesszabo Oct 18 '12 at 9:00
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    $\begingroup$ @ vesszabo:- Emmad has already posted a solution using infinite run of square roots i.e., nested radicals , but where he says that the nested radical is only a close approximation I say that the nested radical is an alternate expression for the left hand side , this can be accomplished by repeatedly substituting the given square-roots in the L.H.S. for 5 $\endgroup$ – Souvik Dey Oct 18 '12 at 9:23

You may be able to apply the logic of Nested Radicals, where:

enter image description here

You may be able to prove that expression:


is a close approximation of:

$$ \sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-5}}}}-5$$

you can show this graphically at least.

you could find the root of $N(x)$ to be $$x=30.$$

Which is also a root for your expression.

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Repeated squaring and simplifying leads to the polynomial equation $$ x^8-204 x^7+18106 x^6-913106 x^5+28616655 x^4-570697702 x^3+7072783751 x^2-49805468751x+152587890630=0. $$ This has eight real roots, including $30$ (and $21$), but none of the other seven roots satisfies the original equation (they satisfy analogues where some of the internal square roots are added rather than subtracted). That's not super satisfying though.

It seems that the function $\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-5}}}}$ is increasing for $x\ge5$; establishing this would prove that $30$ is the only solution. I don't see an easy way to establish it though.

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    $\begingroup$ You (or your computer) was faster than me :-) However the function is, plotting by computer, is not increasing in the interval $(5, 5.004)$. $\endgroup$ – vesszabo Oct 18 '12 at 8:05
  • $\begingroup$ The roots of polynomial (numerically) are $x_1=20.01026, x_2=20.08931, x_3=20.90247, x_4=21\,(exact)\, ,x_5=29.89811, x_6=30\,(exact)\, , x_7=31.00817, x_8=31.09165$. $\endgroup$ – vesszabo Oct 18 '12 at 8:42
  • $\begingroup$ From the original equation we obtain $\sqrt{x-\sqrt{x-5}}=x-(x-25)^2$ which implies that $x-(x-25)^2$ must be positive. It implies $20.4750<x_0 <30.5249$. So only $x_3, x_4, x_5, x_6$ are the possibilities. $\endgroup$ – vesszabo Oct 18 '12 at 8:57
  • $\begingroup$ The function seems to be increasing on the interval $[10,40]$. It would be enough to prove this. $\endgroup$ – vesszabo Oct 18 '12 at 9:07
  • $\begingroup$ @vesszabo: aha, you're right! The derivative is like $-c/\sqrt{x-5}$ just to the right of $x=5$, so it definitely starts out decreasing. $\endgroup$ – Greg Martin Oct 20 '12 at 3:13

Applying the ideas of Souvik Dey(+1) and Emmad Kareem(+1) here is a (hopefully) complete solution.

Assume that $x_0\geq 5$ is a solution of the original equation, that is $$ \sqrt{x_0-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-5}}}}=5. $$ Then from this we obtain $$ x_0-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-5}}}}=x_0-5. $$ Taking the squre root we obtain $$ \sqrt{x_0-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-5}}}}}=\sqrt{x_0-5}. $$ Following this we obtain $$ \sqrt{x_0-\sqrt{x_0- \sqrt{x_0-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-5}}}}}} }} $$ $$ =\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-5}}}}. $$ Here the right-hand-size is $5$. The left-hand-size is a $(4k)$-th subsequence on the (infinite) nested radicals.

So we have to show that $$ \sqrt{x_0-\ldots \sqrt{x_0-\sqrt{x_0-\sqrt{x_0-5}}}}-\sqrt{x_0-\ldots \sqrt{x_0-\sqrt{x_0-\sqrt{x_0}}}} $$ tends to $0$ as the number ($4k$) of square root tends to infinity and applying Emmad Kareem's answer we arrive at the equation $$ N(x)=0. $$ We use the identity $$ \sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}} $$ many times. $$ \sqrt{x_0-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-5}}}}-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0}}}} $$ $$ =\frac{\sqrt{x_0-\sqrt{x_0-\sqrt{x_0}}}-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-5}}}}{\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-5}}}}+\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0}}}}} $$ $$ =\frac{\sqrt{x_0-\sqrt{x_0-\sqrt{x_0}}}-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-5}}}}{5+\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0}}}}} $$ (in the last step we used that $x_0$ is a solution of the original equation) $$ =\frac{\sqrt{x_0-\sqrt{x_0-5}}-\sqrt{x_0-\sqrt{x_0}}}{5+\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-\sqrt{x_0}}}}} $$ $$ \times\,\frac{1}{\sqrt{x_0-\sqrt{x_0-\sqrt{x_0}}}+\sqrt{x_0-\sqrt{x_0-\sqrt{x_0-5}}}}. $$ Here the second fraction can be estimated by $$ \frac{1}{\sqrt{x_0-\sqrt{x_0-\sqrt{x_0}}}}. $$ Since $$ \sqrt{x_0-\ldots \sqrt{x_0-\sqrt{x_0}}}=\frac{1}{2}(-1+\sqrt{1+4x_0})>1 $$ therefore there exists $k_0$ such that this fraction can be estimated by $1$. Using these estimations we obtain the estimation for the $(4k)$-th difference $$ C^{k_0}\frac{5}{5^{k}} $$ which tends to $0$ as $k\to\infty$. (In fact, in our case $\sqrt{x_0-\ldots \sqrt{x_0-\sqrt{x_0}}}\geq\sqrt{x_0-\sqrt{x_0} }\geq 1.$)

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  • $\begingroup$ Yup , this solution is complete and totally correct $\endgroup$ – Souvik Dey Oct 19 '12 at 4:07

Assuming that $X$ is a solution then also $\sqrt{X-5}=5$ and hence $X=30$.

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Observe that $\pm \sqrt{[17,33] + [-7,7]} = \pm [\sqrt{10}, \sqrt{40}] \subset [-7;7]$ This implies that for $x \in [17,33]$, all the quantities $f_1(x), \ldots, f_8(x) = \sqrt{x \pm \sqrt {x \pm \sqrt {x \pm \sqrt {x - 5}}}}$ are well-defined.

Furthermore, for $x=17$, $\sqrt{17 + [-7,7]} = [\sqrt{10}, \sqrt{24}] < 5$, so $f_i(17)<5$
and for $x=33$, $\sqrt{33 + [-7,7]} = [\sqrt{26}, \sqrt{40}] > 5$, so $f_i(33)>5$ .

Thus each of the $8$ equations $f_i(x)=5$ has at least one root in $[17,33]$.

But, a solution of any of these equations is a root of the polynomial $(((5^2-x)^2-x)^2-x)^2-x+5 = 0$, which is of degree $8$, so can have at most $8$ roots.

Reciprocally, if $x$ is a root of this polynomial, then it is a solution of exactly one $f_i(x) = 5$, because the signs of $((5^2-x)^2-x)^2-x, (5^2-x)^2-x, 5^2-x$ determine the signs in front of the square roots from the innermost one to the outermost one.

Therefore, since each of the $8$ equations have at least a solution, they give at least $8$ distinct roots of the polynomial, so there isn't any other root, and thus there is no other solutions to the equations $f_i(x)=5$.

This means that the only solution to $\sqrt{x-\sqrt{x-\sqrt{x-\sqrt{x-5}}}} = 5$ has to be the solution to $\sqrt{x-5} = 5$, which is easily solvable and gives $x = 30$

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