Monday, May 11, 2015

Math Olympiad Problem: Solve for real solutions

Solve for real solutions for the equation $(2x+1)(3x+1)(5x+1)(30x+1)=10$.

Okay, I heard you, why on earth this problem is supposed to be a delicious question that can promote higher mathematics thinking skills?

One of the most common issues math educators are struggling with is the students who underestimate the so-called trivial math problem and they think by the long and typical tedious solving method, the trivial math problem could be safely and successfully solved without a hitch.

True, one can forever adopt the painless and doable tedious method to solve for any given math problem, that is the choice you made, but will it do any good to you? Or one can learn and grow from others so they have all that are required to accomplish anything they choose in life.

Back to the question, what is the doable and tedious method to solve for the given equation $(2x+1)(3x+1)(5x+1)(30x+1)=10$?

That is, we can expand the LHS of the equation, then move the constant to the left and then try to factor the quartic and go from there, like this:







Up to this point, we can divide the equation by $900$ and get:



You could, if you wanted to try the rational roots test to check if there is a way we could find the linear factor(s) for this problem, which would waste us some time to perform this test.

After a few trials, you would agree with me that rational roots test didn't help in this case. So we know we need to factor it into two quadratic expressions, with all $a,b,\,c,\,p,\,q,\,r$ being integers:


If we equate the coefficients of $x^4,\,x^3,\,x^2,\,x$ and the constant respectively from the both sides, we have:






Cool, $5$ variables with $5$ equations, that doesn't seem like a "dangerous" path at all! But that does lead to a computational mess that bog down your progress, often enough to cause you to give up on further solving.

Here is the key to group the "match" factor such that is illustrated below:

If we group the first two factors and the last two factors together, we would not obtain anything useful from the products of these two groups:



But, if we group the smallest and greatest (judging from the coefficient of $x^2$) together and group the two intermediate as another pair, we get:





Ah! If we let $y=60x^2+32x$, we have a rather neat quadratic formula that is fairly easy to solve and back substituting will lead us to the real solutions to the original problem:




$y=-9$  or   $y=4$

In turn, we get:

$-9=60x^2+32x$ when $y=-9$


The discriminant is hence $b^2-4ac=32^2-4(60)(9)=-1136< 0$, which implies $60x^2+32x+9=0$ gives no real solutions.

When $y=4$, we get:

$4=60x^2+32x$ when $y=4$



$x=\dfrac{-8\pm \sqrt{8^2-4(15)(-1)}}{2(15)}$

$\,\,\,\,\,\,=\dfrac{-8\pm \sqrt{124}}{2(15)}$

$\,\,\,\,\,\,=\dfrac{-8\pm 2\sqrt{31}}{2(15)}$

$\,\,\,\,\,\,=\dfrac{-4\pm \sqrt{31}}{15}$

Therefore the only real solutions for the given equation are

$x=\dfrac{-4- \sqrt{31}}{15}$ and $x=\dfrac{-4+\sqrt{31}}{15}$.

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