Is there some way to not use guess and check?

  • M0★ M1★ M2★ M3★ M4 M5

    Module 4 Day 6 Challenge Part 2 Mini-question

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

  • ADMIN M0★ M1 M5

    @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 \( 5 + 7 = \boxed{12}.\)

    🙂