Is there some way to not use guess and check?
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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.
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@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 \( 5 + 7 = \boxed{12}.\)
