Is there some way to not use guess and check?

[The Blade Dancer]

Module 4 Day 6 Challenge Part 2 Mini-question

Is there some way to not use guess and check? Although it is a pretty good strategy, I find it quite unstable.

[debbie]

@The-Blade-Dancer Hi again! This is a good question! Even though guess-and-check sometimes works and can even be the quickest approach sometimes, there's something a bit unsatisfying about it. How do we really know that we'll get anywhere? What if we end up guessing forever? ${\tiny \text{(The horror...) }}$

Well, in this case, we can use a bit of algebra along with the fact that $a$ and $b$ are integers in order to figure out that $a = 7$ and $b= 5.$

Let's first start with the problem statement.

$$\sqrt{a} + \sqrt{b} = \sqrt[3]{22 \sqrt{7} + 26\sqrt{5}}$$

Cube both sides of the equation.

$$\begin{aligned} \left( \sqrt{a} + \sqrt{b} \right)^3 &= \left( \sqrt[3]{22 \sqrt{7} + 26\sqrt{5}} \right)^3 \\ a\sqrt{a} + 3a\sqrt{b} + 3b\sqrt{a} + b\sqrt{b} &= 22\sqrt{7} + 26\sqrt{5} \\ \left( 3b + a \right) \sqrt{a} + \left( 3a + b \right) \sqrt{b} &= 22\sqrt{7} + 26\sqrt{5} \\ \end{aligned}$$

That looks better..... right? (You don't agree it looks better?)

Now let's use the fact that $a$ and $b$ are integers. We know the right hand side has some radicals, the $\sqrt{7}$ and the $\sqrt{5}$ terms. Where could they have come from, though? Could either radical have come from the $\left( 3b + a\right)$ or the $\left( 3a + b \right)$? No, because $a$ and $b$ are both integers, so $\left( 3b + a\right)$ and $\left( 3a + b \right)$ are also both integers. Well, it looks like there's no doubt about where the $\sqrt{5}$ and the $\sqrt{7}$ came from.

 

$$\sqrt{5} \text{ and } \sqrt{7} \text{ came from the } \sqrt{a} \text{ and the } \sqrt{b}$$

So there you have it! Now that you know that $a = 5$ and $b = 7$ (or vice versa... it doesn't actually matter), then you can figure out that their sum is $5 + 7 = \boxed{12}.$