Why let \\(d=0.1\\) through "observation"?

[aaronhma]

In the solution explanation, it said that \(d=0.1\) through "observation" for

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How exactly did we get \(d=0.1\) through "observation" and how do we solve the equation using math instead of "observation"? Thanks.

[quacker88]

@aaronhma That's a great question you bring up! It's awesome that you want to find a more thorough solution to the problem. That's a great learning mindset that you have, keep it up!

By "observation" just means that it was an educated guess. A lot of times, when the numbers are this complicated, it sometimes is faster to just guess and check a few numbers, especially in this case when it's multiple choice. But, I do agree with you-- how would we actually solve the problem?

So, about the problem: one thing I noticed right away is that there actually was a typo in the solution and the decimal should be \(0.5049\)-- I'll get that reported asap.

But besides that, I'll walk you through some of the methods I tried when approaching the problem.

The first thing I saw was just to leave everything unfactored, like so:
\(0.5049=(1-d^2)(1-49d^2)\)
But the problem is, I have no clue how to find two numbers that will multiply to \(0.5049\). If you're clever though, notice how \(5049=99\cdot51\), and plugging in \(d=0.1\) works right away.

So, let's take the expanded version,

\(0.5049 \cdot x = (1-50d^2+49d^4) \cdot x\)
(I set number = \( x\))

After dividing both sides by \(x\) and subtracting both sides by \(0.5049,\) we're left with

\(0.4941-50d^2+49d^4=0\)

Aha! This is a quadratic equation...kinda. If you think about it in terms of \(d^2\), we have a \((d^2)^2\) term, \(d^2\) term, and a constant term, so we can treat it like a quadratic.

To solve a quadratic, we have to factor it! Well...have fun to anyone who tries to do that by hand...
From there I would graph \(0.4941-50d^2+49d^4\) and just find its roots that way (if you're not sure, the roots are when the graph gives a \(y\)-value that is equal to \(0\). in other words, it's when the graph intersects the \(x\)-axis!). Graphing quickly gives you \(d=0.1\) and \(d\approx100.519\) (which is interesting, because there's actually more than one solution! the second one is like ten thousand percent, which is definitely one that no one could have guessed haha)

But, maybe graphing doesn't count as a thorough solution either, since many competitions don't allow calculators. The only thing I can think of now is to just use the answer choices. \(10 \)% is very convenient, so if I were actually taking a multiple choice test no-calculator, I would plug in \(10\)% and hope for the best since it's the simplest answer choice.

Hoping that \(x\)% \( =10\)%\( \implies d=0.1 \) is a solution to the quadratic, we can factor that way. If \(d=0.1\) is a solution, then some version of \((d-0.1)\) must be a factor of the equation (in this case, it's \((0.1-d)\), which is the same thing but multiplied by \(-1\))

Another thing to keep in mind: we're factoring this as a quadratic, so it's going to end up looking something like \((a-d^2)(b-49d^2)\), where the \(d\) are squared. Since 0.1 is a solution to the squared term, then we are hoping that \((0.01-d^2)\) is a factor since \((0.1)^2=0.01\).

And surprise surprise, it works!
Again, kinda cheating since I based a bit of this strategy on the fact that we already know the answer, but here's how the equation looks:

\(0.4941-50d^2+49d^4=0\)
\((0.01-d^2)(49.41-49d^2)=0\)

Since we have the product of two numbers that equal \(0\),
either \((0.01-d^2)=0\) or \((49.41-49d^2)=0\).
solving each case:

\((0.01-d^2)=0\)
\(0.01=d^2\)
\(d=\pm0.1\) (the negative value technically works in this problem too, keep that in mind)

\((49.41-49d^2)=0\)
\(49.41=49d^2\)
\(\frac{49.41}{49}=d^2\)
\(d\approx\pm1.004\)

So the four solutions to \(d\) are \(d=\pm0.1,\pm1.004\). Also, notice how there are four solutions since the highest exponent in the original equation had degree \(4\) (not a coincidence!).
Clearly, this problem is very tedious to do by hand, I definitely wouldn't recommend it. In a problem like this, a lot of the times you just have to guess and check and hope, and also get lucky.

But I hope this helped! It's a lot to digest I know, so if you have any questions let me know