# Handle a Radical Containing a Binomial

I have the following problem : pretty version $\sqrt{x+8} - \sqrt{x-4} = -2$

So far my thinking is to isolate $\sqrt{x+8}$ by changing our equation to $-2 + \sqrt{x-4} = \sqrt{x+8}$

When we square all of it, i'm running into my issue. I don't understand what I should be doing with my binomial left over. At the moment my equation looks like $x + 8 = (-2 + \sqrt{x-4})^2$.

If anyone could show me how to proceed from here it would be much appreciated.

I wouldn't square the equation, at least not in this case. In either way, you need to square two times to get rid of all the radicals.

Solution 1. You could note that $x+8 > x-4$ so the difference of the two radicals cannot be negative.

Solution 2. Note that $\sqrt{x+8}-\sqrt{x-4} = \frac{x+8-x+4}{\sqrt{x+8}+\sqrt{x-4}}=\frac{12}{\sqrt{x+8}+\sqrt{x-4}}$. In this way you can find $\sqrt{x+8}+\sqrt{x-4}$ and by a simple system of equations you can find $\sqrt{x+8}$ and $\sqrt{x-4}$ which in turn will give $x$ right away. Again, in your case you get no solution.

we rearrange $\sqrt{x+8}-\sqrt{x-4}=-2$ to $$\sqrt{x+8}+2=\sqrt{x-4}$$ after squaring we obtain $$\sqrt{x+8}=-4$$ which is impossible.

From the binomial theorem, you know that $(a + b)^2 = a^2 + 2ab + b^2$. Consider the particular case where $a = -2$ and $b = \sqrt{x-4}$. Then the theorem says that

$$(-2 + \sqrt{x-4})^2 = (-2)^2 + 2(-2)\sqrt{x-4} + (\sqrt{x-4})^2.$$

Simplify:

$$(-2)^2 + 2(-2)\sqrt{x-4} + (\sqrt{x-4})^2 = 4 - 4\sqrt{x-4} + |x - 4|.$$

Now if $x < 4$ then $\sqrt{x-4}$ is not a real number. So we can assume $x \geq 4$, so $x - 4 > 0$, and therefore $|x - 4| = x - 4$. So

$$4 - 4\sqrt{x-4} + |x - 4| = 4 - 4\sqrt{x-4} + (x - 4) = x - 4\sqrt{x-4}.$$

$$x + 8 = x - 4\sqrt{x-4}.$$
I think you can see how to simplify this further, but when you do you will find that it is a false equation for every possible value of $x$. Hence the equation you started with was false for every value of $x$.
In fact we can come to this conclusion even easier. The square root is an increasing function, so from $8 > -4$ we know that $x + 8 > x - 4$ and therefore $$\sqrt{x + 8} > \sqrt{x - 4}.$$ That implies that $$\sqrt{x + 8} - \sqrt{x - 4} > 0,$$ that is, it is impossible for $\sqrt{x + 8} - \sqrt{x - 4}$ to be negative, and in particular it is impossible for it to be $-2$. There is no need to square anything and no need for the binomial theorem.
If $x$ is real and you're using the convention that $\sqrt{}$ of a positive number refers to the positive square root, then $\sqrt{x+8} > \sqrt{x-4}$, so it's impossible for the difference to be a negative number.