Trick to this square root equations Okay, so this is a high school level assignment:
$$
\sqrt{x+14}-\sqrt{x+5}=\sqrt{x-2}-\sqrt{x-7}
$$
Here's a similar one:
$$
\sqrt{x}+\sqrt{x-5}=\sqrt{x+7}+\sqrt{x-8}
$$
When solving these traditionaly, I get a polynomial with exponent to the 4th which I cannot solve (I can guess the solutions via free term and divide the polynomial, accordingly, but I don't think that is the intended method here).
Is there a trick to any of these two tasks that would prevent exponents from getting out of control? How would a highschooler solve them?
 A: write like this
$$\sqrt{x+14}+\sqrt{x-7}=\sqrt{x-2}+\sqrt{x+5}$$
square both sides:
$$2x+7+2\sqrt{(x+14)(x-7)}=2x+3+2\sqrt{(x-2)(x+5)}\\
2+\sqrt{(x+14)(x-7)}=\sqrt{(x-2)(x+5)}$$
square again
$$4\sqrt{(x+14)(x-7)}+x^2+7x-94=x^2+3x-10\\
\sqrt{(x+14)(x-7)}=21-x \to (x+14)(x-7)=(21-x)^2$$
Can you finish?
Don't forget to test the result in the original equation.
A: Let's assume for a moment, that those our square roots are actually just that: Square roots of some integer or possibly rational number.
Then, there is the following problem: Sums of square roots are not in general square roots again. For example:
$$x = \sqrt{2} + \sqrt{3}$$
$$\Rightarrow x = \sqrt{5 + 2\sqrt{6}}$$
Both of those forms are not quadratic irrational numbers, and it's highly unlikely you would have to find such an answer in highschool.
But let's look at some examples, which you might encounter:


*

*$\sqrt{3} + \sqrt{12} = \sqrt{27}$: In this case, all the terms share the common factor $\sqrt{3}$ and are integer multiples of that.

*$\sqrt{4} + \sqrt{9} = \sqrt{25}$: In this case, there isn't even such a common square root fator. Actually all terms are integers themselves.


What can we do with that? By simple combination we can figure out the result without doing any much writing at all:
We look closely at the terms under the roots, and see if we can figure out the solution by assuming all of those numbers are actually squares of integers:


*

*We have $x+14, x+5, x-2, x-7$, and all of those are square numbers, because otherwise we could not take the root and still have an integer.

*Clearly if $x-7$ is a square number, then $x \geq 8$.

*Now let's say $x - 7 = a^2$. Then we can rewrite all the other terms as: $a^2+5, a^2+12, a^2+21$.

*If the differences between the terms are small, we can guess that they are in fact neighboring integers. And those are really easy to identify, because the following holds: $(a+1)^2 - a^2 = 2a+1$.


In our terms, these differences are: 5, 7, 9 and that fits quite nicely. In fact, if the difference between $a^2$ and $(a+1)^2$ equals $5$, then immediately we get that $a = 2$. And from $x-7 = a^2$ we get that $x = 11$.
Then finally we would have $2^2, 3^2,4^2$ and $5^2$ under those square numbers. Now all you need to do is to check if that actually makes the equation valid:
$$\sqrt{5^2} - \sqrt{4^2} = \sqrt{3^2} - \sqrt{2^2} (= 1)$$
Regarding the other one: Yes, that can be solved in the same way. The difference between the two lowest terms is $3$, that should already give it away.
