Quadratic FormulaDate: 10/01/1999 at 14:45:00 From: James Panagos Subject: Quadratic Formula Whenever I try using the quadratic formula, I always get it wrong. I need help. Could you give me some ideas on how to do it and some practice problems? Date: 10/06/1999 at 20:06:56 From: Doctor Sandi Subject: Re: Quadratic Formula Okay, James, I'll help you with the quadratic formula. By the time you have used it a few times and over the years, it will be something that you will never ever forget, believe me. The general form of a quadratic equation is y = ax^2 + bx + c, where a, b and c represent constants; e.g. in the equation x^2 + 5x + 4 = 0, the numerical values for a, b and c are a = 1, b = 5 and c = 4. To solve the quadratic equation means to find the values of x that satisfy the equation. There are quite a few methods to solve a quadratic equation, and you have asked about the quadratic formula, so I'll concentrate on this for you. For any quadratic equation of the form ax^2 + bx + c = 0, we can use the following formula to find the values of x (roots) of the quadratic equation: x = (-b +/- sqrt(b^2-4ac))/(2a) where: a = coefficient of x^2 b = coefficient of x c = constant term There will always be two values for x. EVERYTHING in the numerator is divided by the 2a. A very common mistake that students make is to forget this and divide only the numbers under the square root sign by 2a, so watch out for that one. The part of the formula (b^2 - 4ac) that is under the square root sign is called the discriminant. It is usually denoted by a Greek letter that looks like a small triangle. If b^2 - 4ac > 0 then there are two real values for x (that is, the line of the graph touches the x-axis twice). If it is a perfect square, say for example the answer is 81, then we know that the square root of 81 is 9, so if it's a perfect square, it can be evaluated exactly. For all other values, i.e. positive numbers that aren't a perfect square, the factors will contain surds. (Surds - related to the word "absurd" - are irrational numbers that include square roots, cube roots, and nth roots of positive rational numbers.) If b^2 - 4ac = 0 then there are two equal values for x (that is, the line of the graph touches the x-axis once) If b^2 - 4ac < 0 then there will be no real solutions, since we cannot find the square root of a negative number (that is, the line of the graph will not touch the x axis at all). Why am I telling you this? Because sometimes when you have a really complicated function to solve, it is useful to know whether it can be solved at all, or what sort of answer you are going to get. By quickly calculating the discriminant first, you have a good idea what sort of answer to expect from applying the general quadratic formula. For example, as above, if you use the discriminant and the answer you get is less than zero, there is no use going any further, because the quadratic is not going to factor at all. If you solve the discriminant and find that it is greater than 0, then you have actually made your job of solving the quadratic function easier anyway, because you have already found the part under the square root sign. So it does come in handy sometimes. I will give you an example of solving a quadratic equation using the quadratic formula: 2x^2 + 7x -4 = 0 in this case, a = 2 b = 7 c = -4 x = (-b +/- sqrt( b^2 - 4 a c ))/(2 a ) = (-7 +/- sqrt((7)^2 - 4(2)(-4)))/(2(2)) = (-7 +/- sqrt( 49 - 4 (-8) ))/ 4 = (-7 +/- sqrt( 49 + 32) )/ 4 = (-7 +/- sqrt( 81 ) )/ 4 = (-7 +/- 9 )/ 4 = (-7+9)/4 or (-7-9)/4 = 2/4 or -16/4 = 1/2 or -4 So the roots are 1/2 and -4. When you come to using the quadratic equation for modeling, it will probably be the case that one of these answers is not appropriate. For instance, if we are modeling velocity vs. time, we obviously cannot have negative time, so the answer of -4 seconds would be inappropriate and we would discard it in favor of the answer of 1/2 second. But for now you will be finding both answers to get some practice in solving with the quadratic formula. Sometimes you will get questions that look like: Solve the given quadratic equation for x: 2x^2 + 5x = 12. In this case you will need to make the equation equal to zero before you proceed, so you would take the +12 over to the other side, resulting in 2x^2 + 5x - 12 = 0. Note that if, say, the question was 2x^2 - 5x - 12, then a = 2 b = -5 c = -12 When you come to work out the quadratic equation, -b would equal -(-5) = +5, so be careful of that as well. You could also get equations in the form m^2 + 36 = 20m, or a variety of different letters and symbols, in which case you treat them exactly the same as if there were x's, and rearrange in the general form ax^2 + bx + c. In this case, it would be m^2 - 20m + 36, where a = 1, b = -20 (so -b = -(-20) = +20) and c = 36. Do you have a graphics calculator? They are invaluable in helping with the quadratic equations, graphing them, finding maxima and minima, and points of intersection, as well as for differentiation and integration. If you don't have one and you are serious about math, then I strongly suggest that you invest in one. The best thing is to check with your school to see which calculators they endorse - there would be nothing worse than buying a great calculator that did almost everything, only to find that your school won't allow you to use it because it does too many things. That's about all I can tell you at this stage about the quadratic formula. What I will do, though, is give you some equations to solve for x using the quadratic formula, and I'll also give you the answers so that you can check to see if you were right. So, solve the following quadratic functions for x: 1. 2x^2 + 5 x = 12 2. x^2 + 36 = 20x 3. x^2 = 8x - 12 4. x^2 + 4x + 3 = 0 5. x^2 + 4x - 12 = 0 6. 5x^2-6x-7 = 0 7. -3x^2-x-1 = 0 (don't forget that a is represented by -3 in this one so you will be dividing by -6) 8. t^2 - 4t = 32 That should be enough to keep you going for now. The answers are: 1. 3/2 or -4 2. 2 or 18 3. 2 or 6 4. -1 or -3 5. 2 or -6 6. (6 +/- sqrt(176))/10 This is a surd. You can work this out to 1.9266 or -0.7266, but usually answers are preferred in surd form. Actually, (6 +/- sqrt(176))/10 can be further broken down. 176 is actually (16*11) and 16 is a perfect square and can be taken outside the square root as a 4. Canceling everything down, you will end up with (3 +/- 2sqrt(11))/5. This one is a little more complicated, but is the type of question that you can expect once you get a little more into using the quadratic formula. 7. (1 +/- sqrt(-11))/-6 Note here there are no real roots since it is not possible to obtain the square root of a negative number. If you had used the discriminant first to find out whether there were any real roots, the discriminant would have told you that the number under the square root was negative so there are no real roots. 8. -4 or 8 I hope that this helps you. If you have any further questions please send them to Dr. Math. Here are some questions in our archives that deal with answering problems relating to quadratic equations that you might want to look at: Quadratic Trinomials http://mathforum.org/dr.math/problems/rider8.14.html Factoring Polynominals http://mathforum.org/dr.math/problems/factor_polynomials.html Solve by Factoring 3w^2 + 40w - 25 = 20 - 2w^2 http://mathforum.org/dr.math/problems/wfactprob.html Good for you for wanting to learn more. Hope this helped. If you are having problems with any specific questions, please write back. - Doctor Sandi, The Math Forum http://mathforum.org/dr.math/ |
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/