Finding Quadratic Solutions

The topic in my geometry class today involved finding solutions to quadratic equations. The actual topic was on the ratio of lengths of secants in a circle, but that’s for another post.

For a specific example, consider this problem. Two secants intersect outside a circle. The first secant line has segments of 15 (between the intersection point and the circle) and 17 (inside the circle); the second has segments of \(x\) (outside) and 14 (inside). In other words, using the product relationship for secant lines: \[15 \cdot (15 + 17) = x \cdot (x + 14)\]

The next step is to solve for \(x\); the equation can be rewritten to this quadratic equation: \[ x^2 + 14x – 480 = 0 \]

I’m used to using the quadratic formula for such things: \[ x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\] so much so that I can rattle that off rapidly. However, that seems to scare many students.

The standard method for finding solutions for underclassmen, particularly when we reside in the realm of integer solutions, is to find factors of \(c\) that add to \(b\). My concern with numbers like these, though, is that there are a lot of factor combinations to check. 480, for instance, has 24 factors: 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 32, 40, 48, 60, 80, 96, 120, 160, 240, 480. It’s easy to lose track of what you’re doing.

I came up with a quick method for honing in on the correct answer during a prep between classes, but I didn’t have the presentation of it solid enough to present to the students. I’d be curious whether people have tried a method like this, and have opinions on whether it’s successful or not.

In the case that \(c\) is negative, the average of the target solutions has to be close to the square root of \(-c\). So: Start trying integers closest to \(\sqrt{-c} – \frac{b}{2}\). For instance, \(\sqrt{480} – \frac{14}{2} = \sqrt{480} – 7 \approx 14.9\), so try 15. 15 is indeed a factor of 480 (\(480 = 15 \cdot 32\)), but that would give \(b = 17\). This tells us, though, that we’re close; 16 is also a factor of 480, and yields a good pair: \(480 = 16 \cdot 30\), \(30 – 16 = 14\).

One problem I can see is that it only works when \(c\) is negative. If \(c = 480\) and the target factors are 16 and 30, then \(b = 46\) and \(\sqrt{480} – \frac{46}{2} \approx -1.1\), which isn’t even close. Technically, the formula would contain \(\sqrt{-480}\) anyway, but many students would be inclined to ignore the negative sign entirely.

As I said, I do welcome comments. Would this be helpful to students, or just confuse them more?

1 Comment

  1. Pat Ballew

    well for the secant /tangent problems, the constant term will always be negative.
    As far as who did it, you almost duplicated the common Babylonian method of the quadratic formula,
    For your problem x^2 + 14x =480 they would have used sqrt(480 + 7^2) + 7… which gives you one root directly. subtracting the final seven gives the other.

    If you are interested, I wrote about this a long time ago in a paper called “Twenty Ways to solve a quadratic”, you might find some of them interesting. http://www.pballew.net/quadsol.pdf
    Your solution would fall under style number 8.

    Reply

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