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College Algebra: Rational Exponent Equations


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About this Lesson

  • Type: Video Tutorial
  • Length: 7:22
  • Media: Video/mp4
  • Use: Watch Online & Download
  • Access Period: Unrestricted
  • Download: MP4 (iPod compatible)
  • Size: 79 MB
  • Posted: 06/26/2009

This lesson is part of the following series:

College Algebra: Full Course (258 lessons, $198.00)
College Algebra: Equations & Inequalities (50 lessons, $65.34)
College Algebra: Radical Equations (4 lessons, $5.94)

Taught by Professor Edward Burger, this lesson was selected from a broader, comprehensive course, College Algebra. This course and others are available from Thinkwell, Inc. The full course can be found at The full course covers equations and inequalities, relations and functions, polynomial and rational functions, exponential and logarithmic functions, systems of equations, conic sections and a variety of other AP algebra, advanced algebra and Algebra II topics.

Edward Burger, Professor of Mathematics at Williams College, earned his Ph.D. at the University of Texas at Austin, having graduated summa cum laude with distinction in mathematics from Connecticut College.

He has also taught at UT-Austin and the University of Colorado at Boulder, and he served as a fellow at the University of Waterloo in Canada and at Macquarie University in Australia. Prof. Burger has won many awards, including the 2001 Haimo Award for Distinguished Teaching of Mathematics, the 2004 Chauvenet Prize, and the 2006 Lester R. Ford Award, all from the Mathematical Association of America. In 2006, Reader's Digest named him in the "100 Best of America".

Prof. Burger is the author of over 50 articles, videos, and books, including the trade book, Coincidences, Chaos, and All That Math Jazz: Making Light of Weighty Ideas and of the textbook The Heart of Mathematics: An Invitation to Effective Thinking. He also speaks frequently to professional and public audiences, referees professional journals, and publishes articles in leading math journals, including The Journal of Number Theory and American Mathematical Monthly. His areas of specialty include number theory, Diophantine approximation, p-adic analysis, the geometry of numbers, and the theory of continued fractions.

Prof. Burger's unique sense of humor and his teaching expertise combine to make him the ideal presenter of Thinkwell's entertaining and informative video lectures.

About this Author

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Now I want to talk to you about equations that have rational exponents. This is really not a departure from looking at things that have radical exponents because once I did a problem recently that looked like this, , but it could have been easily written this way, (15 - 2x)^1/2 because, remember, a radical is just a power. So, in fact, when you have a rational exponent, which means you just have a fraction in the exponent, solving is not a big deal. It's basically just like having a radical. All we have to do is raise both sides to the appropriate power so that that fractional exponent will cancel out. In this case, I would square both sides. But what if you don't have something. It's just a radical.
Well, let me illustrate this with the following. Suppose I want to solve the following equation; ()^1/4 = x. Now if I want to solve this for x, I could think about this as the fourth root of the whole thing. Well, the method is the exact same that we outlined earlier for radicals which is to first isolate the object that has the funny exponent, the fractional exponent and then just raise both sides to the appropriate power in order to annihilate that. Now in this case what power would I raise it to? Well, you may think, maybe I should square both sides. Well, squaring both sides may help. If I were to square this side, I would then see a power, which means square root. So I'd still have the square root. So let me square again. In particular this tells me that if I look down here, this tells me what power I should use. I should actually raise both sides to the fourth power. So if I fourth power both sides, so power up this equation if you will--a little power joke there. It wasn't a very powerful joke but that was cute. They can't all be gems. If I raise both sides to the fourth power, what do you see? Well, on this side I have something to the power all raised to the fourth power. The laws of exponents tell me what I do is multiply these two numbers so x 4 is just one. So these guys actually annihilate themselves, which was the whole point of doing this. And so all I'm left now is with the inside. 5x^2 - 6 and that equals x^4.
So now I've reduced this equation, which had a fourth power exponent or a fourth root into this thing. Now that looks like a fourth degree equation so it looks pretty scary, but let's not panic quite yet. There's plenty time to panic later. Let's now pull everything over to the one side. Now usually I pull things over to this side, but if you see I pull it over to that side, I would have a -x to the fourth which actually sort of bothers me because I like to have the highest power always positive. So if you will indulge me, let me bring everything over to this side now. If I bring everything over this side, just so that this will always remain positive looking, I would have 0. It also illustrates that it doesn't make a difference. You can bring them over side. If you're right handed, left handed, math is open for everybody. This would be x^4. If I bring this over now, I'm subtracting so that would be a -5x^2 and if I bring this over, I have to add 6 to both sides so I'd see a +6. Well, now you're saying, Gee whiz. Look at that. That's a fourth degree. How am I going to handle that? Well, the way I'm going to handle that is notice that I can make a substitution. I can write this as x^4 is just x^2 and everything else stays the same. But now you can see that really this is just a quadratic equation, you see? And x squares. So what I could do is make a substitution. If you like making substitutions, this is a good time to do it. So I'm going to say that let's let a = x^2. So if a = x^2, then I can replace this thing by an a and that thing by an a. And what would it look like? If I do that, it would look like this. Now I'd see the following. I would see, well, a^2 - 5a +6 and that still equals 0. Well, let's now factor this. I hope it factors. I have a and a. A positive sign means they're both the same sign and they're both negative. Multiply to give 6 and add to give 5 so that looks like a 2 and 3. So if (a - 2) (a - 3) = 0, and so what does that mean? It means that a = 2 or a = 3. So there are two solutions to this, but that's not x. That's just a. I've got to now remember that a = x^2. So, in fact, what I can say now, what I can say is the following. I can say that either x^2 = 2 or x^2 = 3. And now how do I solve each of those equations separately? Well, let's take a look at each one. I can just take plus or minus the square root. That may be the easiest thing to do so I'd see here if x^2 = 2, that means that . There are two solutions there. And if x^2 on the other hand equals 3, that means that . So actually we see four possible answers now. Right? I see or or or . I have to check each and every one of them back in here. So let's do that right now and see how many we can do. Maybe I'll do a couple of them and let you try the rest on your own.
So let's try the first one. If I try, let's say the . So let's check . Let's see what happens. So I plug in the whenever I see an x. So I see 5 times--now the squared is just 2. So that's 5 x 2 - 6 to the power and does that equal the square root of 2. Well let's see. 5 x 2 is 10 and 10 - 6 is 4. So I see 4^1/4. Does that equal the square root of 2? Well this actually may look like it's not equal, but let's think about what the power means. It means, well, one way of thinking about it would be the following. I could say it's just 4^1/2 x 1/2 because that's just . And , after all, means square root. So this just equals the square root of 4 all to the power. The exponent means square root. The square root of 4 is 2 so this is just 2 and the means; this thing means square root. So, in fact, what I see is . This answer actually checks. If you check that over carefully and see that in fact, the checks. I'll let you try the other ones in the quizzes, but let me just tell you that will not check and I will let you figure out which of these solutions or actually works. But try it. You have to check all your answers. So this problem is still incomplete, but you'll finish it up.
Equations and Inequalities
Radical Equations
Solving an Equation With Rational Exponents Page [1 of 2]

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