$\sqrt{3-x}-\sqrt{-x^2+8x+10}=1$

Question

Consider the equation $$ \sqrt{3-x}-\sqrt{-x^2+8x+10}=1. $$

I have solved it in a dumb way by solving the equation of degree four. So, the only real solution is $x = -1$. Can you please suggest, maybe there are better or easier ways to solve it?

Answer 1

This is a simple way used when I was in high school. Basically, all you have to do is draw out the minimal polynomial of the known solutions (which make this even useful with most algebraic solutions, not only integers).

Firstly, match each square root with its real value.

\begin{align} (\sqrt{3-x}-2)-(\sqrt{-x^2+8x+10}-1)=0 \end{align} Then multiply each with its conjugate.

\begin{align} \frac{-1-x}{\sqrt{3-x}+2}-\frac{-x^2+8x+9}{\sqrt{-x^2+8x+10}+1}=0 \end{align} \begin{align} (1+x)\left (\frac{1}{\sqrt{3-x}+2}+\frac{9-x}{\sqrt{-x^2+8x+10}+1} \right )=0 \end{align} From the fact that $3-x>0$ and $-x^2+8x+10>0$, it can easily be checked that $9-x>0$ Therefore, the only solution would be $x=-1$.

Answer 2

Note that the two square root functions must be non-negative, that is,

$$3-x \geq 0 \Rightarrow x \leq 3$$ $$-x^2+8x+10 \Rightarrow 4 - \sqrt{26} \leq x \leq 4 + \sqrt{26}$$

Upon intersecting the two inequalities, we have $4 - \sqrt{26} \leq x \leq 3$. You can try integers in the inequality.

Answer 3

You did it in the right way. You will end up with a $x^4$ term, and you must solve from there to get $-1.$

The above answer is right, that it gives you a $range$ of values, but if you want to solve directly, you must do it the way you have said above.