How To Really Solve This Tricky Algebra Problem? (VI)

Welcome to the sixth entry in the tricky algebra problem series. Following from the last entry, we are still lingering on the secondary-math-Olympiad level. This time, you are presented with two equations featuring two unknowns (x and y):

x + y = 41

x*y = 16

The first equation features an addition operation involving the two unknowns, and the second equation features a multiplication operation featuring the two unknowns.

Given this premise, your challenge is to figure out the numeric solution to the following challenging-looking equation:

x √(x) + y √(y) = ??

Do you think you can solve this?

Hint:

You can start solving this problem by using the binomial square formula.

Spoiler Alert:

Beyond this section, I will be explicitly discussing solutions to the problem. So, if you wish to solve this problem on your own, now is the time.

After you’ve completed your attempt, you may choose to continue reading the essay to compare your approach with mine.

This essay is supported by Generatebg

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Applying Logic to the Given Equations

The first and foremost thing that I’d like to do is to number the given equations as equation 1 and equation 2 respectively. This will make it convenient to refer to them later.

After this, let us focus on equation 2. It features a multiplication operation between the two unknowns and yields a positive result on the right-hand side. This means that either both the unknowns are positive numbers or both are negative numbers.

With this insight in mind, let us now shift our focus toward equation 1. It features an addition operation involving the two unknowns and yields a positive result on the right-hand side. If the two unknowns were both negative numbers, this would not be possible.

So, by logical exclusion, we could conclude that both the unknowns are positive numbers:

x > 0 and y > 0

Applying the Binomial Square Formula to the Tricky Algebra Problem

At the beginning of the essay, I hinted that you could start solving this problem using the binomial square formula. We are about to do this now. The trick is to notice that the final equation that we are trying to solve features terms that are square roots of the two unknowns (√(x) and √(y)).

Let us try and apply the binomial square formula to the square of the sum of the square roots of both the unknowns (boy, that was a mouthful!). It is probably far easier to take a look at how we can work this equation:

We have conveniently arrived at a numeric result on the right-hand side. Now, we could just take the square-root on both sides as follows:

There we go. For now, let us register this equation as equation 3. But where do we go from now?

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Applying the Binomial Cube Formula to the Tricky Algebra Problem

This sounds counter-intuitive, but think about it. The final equation that we are solving for not only features square-root terms of our unknowns, but it specifically features ‘x√(x)’ and ‘y√(y)’.

We could arrive at these terms by using the binomial cube formula. If this is not clear yet, it will become clear in a moment:

We are now on the home-stretch; just a couple of steps more.

How to Solve the Tricky Algebra Problem

When we plug in the numeric values we already known from equation 2 and equation 3, we end up solving the following equation:

There we go. The solution to our problem is 259.

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Further reading that might interest you: How To Really Solve The Kissing Circles Puzzle? and How Are Big Airships Really Making A Comeback Again?

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