Duplicate Disclaimer: At the bottom

This question is mainly about heuristics; I would like to learn how to think when doing these problems, and I am just using $\sqrt{a-\sqrt{a+x}}=x$ as a case study.

I am very interested in becoming better at algebraic manupulation; but many times I get stuck and don't know what to do. There are many many ways to go forward which all look equally good to me but I don't know which to choose, and how far to go on each way.

As an illustration I will try to solve $\sqrt{a-\sqrt{a+x}}=x$

I can easily transform this into $(x^2-a)^2=x+a$. At this point I try different things:


This yields $x^4-2ax^2-x+(a^2-a)=0$. Since factors must multiply to the constant term, I try $a$ and $1-a$ but they don't work. At this point I recall that it is possible to factor a quartic by adding and subtracting a clever value, and then factoring by grouping. But I don't know which terms to group and how many terms should be in each group. At this point I start wondering if expanding was bad and I shouldn't have thrown away the nice factored form.


Back at $(x^2-a)^2=x+a$. There is nothing obvious to do (barring expanding) so I try substitutions:

$1)$ Eliminate the $a$ on the left by $x=y+\sqrt a$ or by $x^2=y^2+a$. This yields $(y^2+2 y\sqrt a)^2=y+a+\sqrt a$ or $y^4=\sqrt{y^2+a}+a$ respectively, which don't seem any nicer.

$2)$ Eleminate the $a$ on the right by $x=y-a$, which doesn't lead to anything nice.

$3)$ Scale: $x=ay$, or $x=\sqrt ay$ which again doesn't help.

$3)$ Translate and scale $x=ay - \sqrt a$ which doesn't lead to anything nicer either.

I could try to go farther on each of these ways, but I never know if I should stop because it's a dead end.

How can I improve my ability to solve algebraic manipulation problems?

Duplicate Disclaimer: This is not a duplicate of this because answers just stating the solution would suffice for that question. Such answers would not suffice for my question, because I am interested in an explanation of the heuristics involved in solving such questions in general.


Sugestion: Try to replace the role of variables. Solve the equation on $a$ first. That aproach helps many times.

From $x^4-2ax^2-x+(a^2-a)=0$ we have:

$$ a^2-a(2x^2+1)+x^4-x=0$$ so we have $$ a_{1,2} = {2x^2+1\pm (2x+1)\over 2}$$

  • $\begingroup$ and from where do you know, that this "Solutions" fulfill your equation? any conditions on the parameter? $\endgroup$ – Dr. Sonnhard Graubner Dec 12 '17 at 18:11

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