Solving Rational, Radical, and Polynomial InequalitiesDate: 12/28/2003 at 20:47:57 From: Chapmann Wong Subject: How do I solve inequalties algebraically How do I solve rational and radical inequalities algebraically? I understand how to solve a rational inequality using a number line. For example; (x - 2) ------- > 0 (x + 4) The number line is split into 3 parts: x < -4, -4 < x < 2, x > 2. A "test point" number from each of the sections is taken and plugged into the original inequality, and if it makes the inequality correct, then the whole section is a solution to the inequality. The solutions for this inequality are x < -4, x > 2. But when I use algebra, all I get is x > 2. Why is this? Is there something about rational inequalities that make them have to be solved with a number line? I also have trouble with radical inequalities. Is there something with the square root that makes it not possible to solve by algebra? Date: 12/29/2003 at 07:21:52 From: Doctor Rick Subject: Re: How do I solve inequalties algebraically Hi, Chapmann. Thanks for writing. First, the short answer--you should be able to solve any inequality you run into using algebra. But there are a few ideas to keep in mind. When solving your sample problem, you need to consider separate cases for whether the denominator is positive or negative. If x + 4 > 0 (positive), then when you multiply both sides by x + 4, you don't change the direction of the inequality. There are now two conditions on x: x + 4 > 0 AND x - 2 > 0 If x + 4 < 0 (negative), then when you multiply both sides by x + 4 you need to change the direction of the inequality. The two statements for this case are: x + 4 < 0 AND x - 2 < 0 The two cases (positive or negative) lead to x > -4 AND x > 2 OR x < -4 AND x < 2 The first of these cases reduces to x > 2, since any number greater than 2 is also necessarily greater than -4. The second case reduces to x < -4, since any number less than -4 is also necessarily less than 2. Thus the algebraic solution is: x > 2 OR x < -4 Do you see where you went wrong? Algebra does work, but you need to consider cases in a manner similar to the number-line method. Any time you multiply through by something which could be positive or negative, you need to create a separate case for each possibility. If you can't see yet how to correct your algebraic approach to inequalities with radicals, show me an example of such a problem and I'll go through it with you. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 12/29/2003 at 11:39:07 From: Chapmann Wong Subject: How do I solve inequalties algebraically Thank you for the information. When trying to solve the following problem (x - 4) ------- >= -2 (x + 2) I again got the wrong answer. I tried making cases but it didn't seem to work. The correct answer is x < -2 or x >= 0. Where did the zero come from? Thank you! Date: 12/29/2003 at 12:26:08 From: Doctor Rick Subject: Re: How do I solve inequalties algebraically Hi again, Chapmann. With this new problem, you need to multiply both sides by (x + 2), and again you need to keep track of whether the multiplier is positive (x + 2 > 0) or negative (x + 2 < 0), so there are two cases: x + 2 > 0 AND x - 4 >= -2(x + 2) OR x + 2 < 0 AND x - 4 <= -2(x + 2) x > -2 AND x - 4 >= -2x -4 OR x < -2 AND x - 4 <= -2x - 4 x > -2 AND 3x >= 0 OR x < -2 AND 3x <= 0 x > -2 AND x >= 0 OR x < -2 AND x <= 0 The first case reduces to x >= 0; the second reduces to x < -2. Again, the key idea is to consider what happens if the multiplier is positive and what happens if it's negative. The two conditions in each case are the multiplier and the result of the inequality after multiplying through. If you don't follow any of these steps, please ask about it. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 12/29/2003 at 14:35:12 From: Chapmann Wong Subject: How do I solve inequalties algebraically Dr. Rick - Thank your for all your help. I think I'm getting the rational inequalities now. But I'm still a little confused about radical inequalities. Can you help me with those? Here's an example, and what I did on it: _____ \/x + 2 >= x I squared both sides and got the following possibilities: x + 2 >= x^2 so then 0 >= x^2 - x - 2 so then 0 >= (x - 2)(x + 1) OR x + 2 <= x^2 so then 0 <= x^2 - x - 2 so then 0 <= (x - 2)(x + 1) So I took each of the two cases and solved them, and overall I came up with -1 <= x OR x >= 2. However, the answer is -2 <= x <= 2. Why is this? I didn't get a -2 anywhere. Or did I do something wrong in the process? Thanks again for all your help. Date: 12/29/2003 at 15:21:52 From: Doctor Rick Subject: Re: How do I solve inequalties algebraically Hi Chapmann - Problems with radicals take some thinking; you always need to check your results to see if they make sense. When you square both sides of an equation or inequality, funny things can happen. For example, here's a true statement: -2 < 2 But if you square both sides, you get a false statement: 4 < 4 Similarly, you can turn a false statement into a true one by squaring: -3 = 3 (False) 9 = 9 (True after squaring both sides!) So we tend to approach radical problems the same way we always do, but keep it in your mind that you need to really think about and check the answer you get. There's one other important idea to keep in mind when working with radicals--you can't take the square root of a negative number if you are working with real numbers only. So hold onto that thought, too. I like your idea of squaring both sides, but I don't agree that you would get two cases as you show. In fact, if you look at your two cases 0 >= (x - 2)(x + 1) OR 0 <= (x - 2)(x + 1) closely, you'll see that you are saying that the product of (x - 2) and (x + 1) can be negative (case 1), positive (case 2) or zero (both cases). In other words, it can be anything! So the answer to your approach would be that x can be any real number, which is clearly not the correct solution to the problem! Squaring both sides just produces one case, and it does not change the inequality (though we need to check our answers, remember), so the correct case is your first one 0 >= (x - 2)(x + 1) In order for the product to be negative, one factor must be positive and the other negative. In other words, one of these is true: x - 2 <= 0 AND x + 1 >= 0 OR x - 2 >= 0 AND x + 1 <= 0 x <= 2 AND x >= -1 OR x >= 2 AND x <= -1 -1 <= x <= 2 OR impossible Thus the apparent solution is x between -1 and 2. However, when we look back at the original problem, there are two considerations. 1. The result of the square root can never be negative since the way it is written refers to the principal or positive root. Thus, if x is negative, the inequality must be true (as long as the radical makes sense) since a positive will be greater than any negative. 2. The square root only makes sense for positive numbers (if we're talking about real numbers, not complex numbers). Thus the equation only makes sense if x + 2 >= 0, that is, if x >= -2. For positive numbers x, we know x must be less than or equal to 2. For negative numbers, we have now observed that the equation will be true for any negative x that is allowed, namely x greater than or equal to -2. The correct solution therefore is -2 <= x <= 2. The thing to remember is that whenever you square both sides, you are likely to introduce spurious solutions, so you must check your answer and adjust the solution set as needed. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 12/29/2003 at 16:02:12 From: Chapmann Wong Subject: How do I solve inequalties algebraically Hi Doctor Rick, Thanks so much for the information that you've given me. It has helped so much. By the way, do I follow the same format for solving polynomial inequalities such as (x - 2)(x + 1)^2(x - 4) <= 0? Or is there a different method? Date: 12/29/2003 at 17:25:31 From: Doctor Rick Subject: Re: How do I solve inequalties algebraically Hi, Chapmann. In general for polynomial inequalities you should set one side to 0 and factor the other side completely. Your problem is already in that condition. There are several methods that could be used, or combined in various ways. One would be to do a number line analysis as you mentioned at the beginning, with critical values at -1, 2 and 4. Another would be to make a series of statements realizing that for the result to be negative (< 0) you would need either one negative factor and two positive factors or three negative factors. This task is reduced slightly if you realize that (x + 1)^2 will always be positive, but it's still a lot of statements and is not recommended. My first thought when I see your example is to think about the graph of the polynomial function y = (x - 2)(x + 1)^2(x - 4). The polynomial has four roots: -1 (double root), 2, and 4. I can sketch the graph. It is positive to the right of x = 4, since the coefficient of the highest power is positive. From right to left, it crosses the x axis to the negative side (below the x-axis) at x = 4, recrosses to the positive side at x = 2, then touches the x axis on the positive side at x = -1 before rising again. Thus I can see that it is negative in 2 < x < 4, and zero at each of the roots. The solution is 2 <= x <= 4, x = -1 This method is quite similar to your number line analysis, and as you learn more about polynomial functions and their graphs, you'll see why they are related. Good luck! - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 01/04/2004 at 12:48:08 From: Chapmann Wong Subject: Thank you (How do I solve inequalties algebraically) Thank you for helping me learn more about math. It has helped me greatly. Once again, thank you very much! |
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/