This post is the first of a series of hints and techniques for students of math, all based on my experience as a math tutor and teacher…

Much of what a student must learn in a math course amounts to mastering the steps in a well defined procedure. For example, multiplying two binomials (e.g. (x-3)(y²+z) ) requires a fixed set of simple steps, and is easily mastered. Same for factoring trinomials, solving quadratic equations, and dozens of other skills. More than memorization is needed, of course, but remembering the steps and/or formulas is an essential part of the learning process.

Not so with word problems – with word problems, one must somehow understand what the problem is saying and what it is asking us for, then we must convert all those words to mathematical language (usually, to an equation). It’s much harder for a teacher, tutor, or textbook author to lay out the steps for doing all this. No wonder that students often say that they don’t like word problems.

In this article, I will describe a method applicable to many word problems, a method I call “formula driven problem solving”. Simple as it is, I have seen it help students solve problems that they would otherwise have starred at, wondering where to start.

By way of illustration, consider this little problem, which could easily be on the SAT:

Find the area of this isosceles triangle in terms of p:

Once we recall that the common formula for triangle area is ½ x base x height, and the base is p, it’s obvious that we must find the height:

The shaded inner triangle is a right triangle with sides p/2, h, and 2p. By the Pythagorean Theorem,

\[ \left(\frac{p}{2} \right)^{2}+h^{2}=\left(2p\right)^{2}\]

\[ h^{2}=4p^{2}-\frac{p^{2}}{4}=\frac{15}{4}p^{2}\]

so

\[ h=\frac{\sqrt[]{15}p}{2}\]

And

\[ area=\frac{1}{2}p\frac{\sqrt[]{15}p}{2}=\frac{\sqrt[]{15}p^{2}}{4}\]

This problem may seem trivial to some, but I know from experience that many students would fail to get started because they don’t say to themselves “we want the area of a triangle, so the solution must involve area = ½ x base x height”.

Here’s another geometry-based example:

A circle has circumference 12. What is its area?

The first step is to think “the problem wants the area of a circle. That formula is
\[a = \pi r^2\]
So I have to somehow find r, the radius”.

Next, we say to our self “I’m given the circumference. How can I find the radius, given the circumference?” Now we remember another formula:
\[C=\pi d=2\pi r\],
where d is the diameter and r is the radius.

Therefore, we can write

\[r=\frac{C}{2\pi}=\frac{12}{2\pi}=\frac{6}{\pi}\]

So then we have it:
\[area=\pi\,\left(\frac{6}{\pi}\right)^{2}=\frac{36}{\pi}\]

Now a third and final example:

A straight line has slope ¾, and passes through the point (4,3). What is the equation of the line?

Here the thought process would be “The general formula for the equation of the line is y = mx + b, where m is the slope, and b is the y intercept. We know the slope is ¾, so y = ¾ x + b, so I must find b. Evidently, that must come from the fact that the line passes through the point x = 4, and y = 3″.

Thus, 3 = ¾ · 4 + b

From which b = 0

Therefore, y = ¾ x is the equation.

Finally, here is a practice problem:

Find the simplest quadratic equation whose roots are

\[\frac{1\pm\sqrt[]{5}}{2}\]

(the formula in this case is, of course, the quadratic formula):

\[ x=\frac{-b\pm\sqrt[]{b^{2}-4ac}}{2a}\]

Ans: x² – x – 1 = 0

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This entry was posted on Thursday, July 23rd, 2009 at 11:19 am and is filed under Learning Math, Problem Solving Techniques. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.