Factoring Quadratics Without GuessingDate: 05/28/2002 at 16:41:09 From: Chris Subject: Quadratic Equations Hi. My name is Chris and I am a seventh grader, but I love math and like to learn on my own. I have already learned basic trigonometry, logs, etc... I am now learning how to solve quadratic equations. I found out how to, but I am still wondering how to do one thing. It is part of the factoring. I will use the example x^2 + 5x + 6 = 0. My dad taught me that you find two factors of the 6 that add up to the 5 in 5x. In this case it can be 2 and 3. He told me to then make it into (x + 3)(x + 2) = 0. What I am wondering is if there is any way to find two factors of 6 (or whatever number it is) that add up to 5 (or the other number) without using guess and check. If there is no way to do that, then how do you factor it when the two factors are not whole numbers? I would love to know this. -Chris Date: 05/28/2002 at 17:54:57 From: Doctor Ian Subject: Re: Quadratic Equasions Hi Chris, In fact, guess and check is the most straightforward way to proceed if you have reason to believe that the roots are integers, and what makes it reasonable is that there usually aren't all that many possible guesses! This is because for any given integer, there are a limited number of ways to multiply two other integers to get it. For example, consider something like x^2 + 13x - 68 If you break 68 into prime factors, 68 = 2 * 2 * 17 it becomes clear that the _only_ possible ways to factor -68 are -1 * 68 1 * -68 -2 * 34 2 * -34 -4 * 17 4 * -17 and only one of these pairs of factors can add up to 13. So it's guess and check, but when done correctly, it's a very _educated_ kind of guess and check. But there _is_ a way of avoiding guess and check, called 'completing the square', which makes use of two very common patterns: 1. (x + a)^2 = x^2 + 2ax + a^2 2. (x + a)(x - a) = x^2 - a^2 You're going to be seeing a lot of these as you continue on in math, so you may as well take this opportunity to become friends with them. :^D How do we use them? First, we complete the square by adding the square of half the coefficient of x to both sides of the equation. Using your example equation, x^2 + 5x + 6 = 0 x^2 + 5x + (5/2)^2 + 6 = (5/2)^2 (x + (5/2))^2 + 6 = 25/4 Now we subtract it from both sides to get (x + (5/2))^2 + 6 - 25/4 = 0 (x + (5/2))^2 + 24/4 - 25/4 = 0 (x + (5/2))^2 - 1/4 = 0 And since 1/4 is a square (of 1/2), we can use our second pattern: (x + (5/2))^2 - 1/4 = 0 [(x + 5/2) + 1/2][(x + 5/2) - 1/2] = 0 [x + 3][x + 2] = 0 It has a certain elegance, doesn't it? But as you can see, the numbers can get a little hairy, which is why people tend to prefer the guess and check method. It offers fewer chances for making silly mistakes. But the quickest way to factor a quadratic equation is to use the quadratic formula. Again, using your example, x^2 + 5x + 6 = 0 the quadratic formula says that the roots are -b +/- sqrt(b^2 - 4ac) x = ---------------------- here a=1, b=5, c=6 2a -5 +/- sqrt(5^2 - 4*1*6) = ------------------------ 2*1 -5 +/- sqrt(25 - 24) = -------------------- 2 -5 +/- 1 = -------- 2 = -4/2, -6/2 = -2, -3 This tells us that the original equation is zero when x = -2 or when x = -3; so x^2 + 5x + 6 = (x + 2)(x + 3) Note that you have to be careful with signs when using this method! It's easy to get them mixed up. The quadratic formula is by far the fastest way to factor a quadratic equation, and it has the advantage that it works even when the roots aren't integers. In practice, I'll usually give factoring about 5-10 seconds, and if I don't get it immediately, I'll go right to the quadratic formula. And when the initial coefficient is anything other than 1, e.g., 3x^2 - 2x + 15 I won't even _try_ factoring, because the number of possibilities goes up so fast that I get a headache just thinking about it. Note that you can use the 'discriminant', sqrt(b^2 - 4ac) to quickly predict whether you're going to end up with something neat, or something messy. As we saw, for x^2 + 5x + 6, you get sqrt(5^2 - 4(1)(6)) = sqrt(25 - 24) = 1 which is pretty tame. For x^2 + 13x - 68, we get sqrt(13^2 - 4(1)(-68)) = sqrt(169 + 272) = 21 which is also pretty tame. For x^2 + 13x + 13, we get sqrt(13^2 - 4(1)(13)) = sqrt(169 - 52) = sqrt(117) which is somewhat ugly. And for x^2 + 13x + 68, we get sqrt(13^2 - 4(1)(68)) = sqrt(169 - 272) = sqrt(-103) which means that the roots are complex instead of real: http://mathforum.org/library/drmath/view/53877.html Don't worry too much about complex numbers right now. I just wanted to include them here because you've probably already realized that a quadratic equation describes a polynomial, and that the roots of the polynomial tell you where the parabola crosses the x-axis. Eventually it would dawn on you that it's possible to draw a parabola that doesn't cross the x-axis... which would lead you to wonder: Doesn't the equation of that parabola have roots? And the answer to that question is: It has roots, but they are complex instead of real. I hope this helps! Good luck with your studies, and don't hesitate to write back if you'd like to talk more about this, or anything else. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 05/28/2002 at 20:55:36 From: Chris Subject: Thank you Thank you so much for answering my question. It helped me a lot. Thanks. -Chris |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/