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What are the general ways to solve radical equations similar to questions like

$\sqrt{x+1}+\sqrt{x-1}-\sqrt{x^2 -1}=x$

$\sqrt{3x-1}+\sqrt{5x-3}+\sqrt{x-1}=2\sqrt2$

$\sqrt{\frac{4x+1}{x+3}}-\sqrt{\frac{x-2}{x+3}}=1$

Are there just a few known ways to solve them? How do you know the best way to solve such questions? I have trouble with a lot of square root equations, and when I ask them on this site, I get good answers, but for one question. I was wondering if there were any general principles of solving such questions.

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    $\begingroup$ The most basic is to isolate one of the square roots, square both sides, and this eill usually yield less square roots than when you started with. Keep repeating until you have no more square roots. This works pretty well in most cases, but as you can see by Dr. Graubner's answer, there are sometimes some tricks, arising from the particular problems themselves ways, which allow you to solve the equations much faster $\endgroup$ – Ovi May 21 '16 at 0:07
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    $\begingroup$ Example question (with answers on how to approach it) $\endgroup$ – ccorn May 25 '16 at 3:41
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One rather general strategy is to replace each new root $\sqrt[k]{expression}$ in the equation by a new variable, $r_j$, together with a new equation $r_j^k = expression$ (so now you will have $m+1$ polynomial equations in $m+1$ unknowns, where $m$ is the number of roots). Then eliminate variables from the system, ending with a single polynomial equation in one unknown, such that your original variable can be expressed in terms of the roots of this polynomial. This procedure can introduce spurious solutions if you only want the principal branch of the $k$'th root, so don't forget to check whether the solutions you get are valid.

For example, in your second equation, we get the system $$ \eqalign{r_1 + r_2 + r_3 - 2 \sqrt{2} &= 0\cr r_1^2-(3x-1) &= 0\cr r_2^2-(5x-3) &= 0\cr r_3^2-(x-1) &=0\cr}$$ Take the resultant of the first two polynomials with respect to $r_1$, then the resultant of this and the third with respect to $r_2$, and the resultant of this and the fourth with respect to $r_3$. We get $$ 121 x^4-4820 x^3+28646 x^2-45364 x+21417$$ which happens to factor as $$ \left( x-1 \right) \left( x-33 \right) \left( 121\,{x}^{2}-706\,x+ 649 \right) $$ However, only the solution $x=1$ turns out to satisfy the original equation.

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  • $\begingroup$ also with your method we have to solve a polynomial, can you factorize this thing by hand? $\endgroup$ – Dr. Sonnhard Graubner May 21 '16 at 0:32
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    $\begingroup$ I confess that I did not do this by hand, though in principle the Rational Roots Theorem might be helpful to one trying such a foolish thing. $\endgroup$ – Robert Israel May 21 '16 at 0:40
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    $\begingroup$ No, you use a Computer Algebra system. $\endgroup$ – Robert Israel May 21 '16 at 0:43
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    $\begingroup$ Humans also make mistakes. Computers can and do prove things. $\endgroup$ – Robert Israel May 21 '16 at 0:50
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    $\begingroup$ Most calculators would not be terribly useful for this. $\endgroup$ – Robert Israel May 23 '16 at 0:56
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I would write the equation in the form $$\sqrt{x+1}+\sqrt{x-1}=x+\sqrt{x^2-1}$$ with $$x\geq 1$$ after squaring one times and isolating the square root we get $$2\sqrt{1-x^2}(1-x)=2x^2-2x-1$$ squaring again we obtain $$(x^2-1)(2-2x)^2=(2x^2-2x-1)^2$$ this gives the equation $$4x-5=0$$ and we get $$x=\frac{5}{4}$$ fulfills our equation. in the second equation we get $$x=1$$ write your third equation in the form $$\sqrt{\frac{4x+1}{x+3}}=1+\sqrt{\frac{x-2}{x+3}}$$ we get after squaring two times $$\left(\frac{2x}{x+3}\right)^2-4\left(\frac{x-2}{x+3}\right)=0$$ after simplifying we obtain$$-4\,{\frac {x-6}{ \left( x+3 \right) ^{2}}}=0$$

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    $\begingroup$ Thanks; I'm not asking for specific answers though, I'm asking for a more general idea of how to solve similar questions. $\endgroup$ – suomynonA May 20 '16 at 23:57
  • $\begingroup$ i have read you question carefully but i think such a general procedure doesn't exist we must try to get a polynomial in $x$ which we can solve or we take a numerical method $\endgroup$ – Dr. Sonnhard Graubner May 21 '16 at 0:00
  • $\begingroup$ For example, one way is replacing the roots with variables; are there are any other ways? $\endgroup$ – suomynonA May 21 '16 at 0:04
  • $\begingroup$ yes this is one way but then we get often a non linear equation system which is in the general not easy to solve $\endgroup$ – Dr. Sonnhard Graubner May 21 '16 at 0:07
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    $\begingroup$ Ok, but my question still stands: How do you identify and solve questions of such "special form"? $\endgroup$ – suomynonA May 21 '16 at 20:47

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