Square Root of a Negative Number SquaredDate: 01/23/97 at 12:05:48 From: Cindy Smith Subject: Re: Exponent Rules Hi, I'm reading popular math books in addition to my textbook and am really trying to understand the material. Here's my question: I understand that the nth root of x^n = x^(n/n) or x^1. My textbook gives two problems: Sqrt(-6)^2 and the 4th root of (-3)^4. This leads to two different solutions in my mind. (-6)^2 = 36 so the square root of (-6)^2 = 6. Similarly, the 4th root of (-3)^4 leads to the fourth root of 81, which is 3. However, according to the rules, the nth root of x^n = x^(n/n). Therefore, the square root of (-6)^2 should equal (-6)^(2/2) or (-6)^1, which equals -6. By a similar method, the fourth root of (-3)^4 = (-3)^(4/4) or (-3)^1, which equals -3. The answers in the back of the book says the correct answers are positive 6 and 3. Where is the error in my logic? Thanks again for all your help! Cindy Smith Date: 01/23/97 at 20:28:04 From: Doctor Keith Subject: Re: Exponent Rules Great question! Please note that all questions are welcome, so you should write as many as you like. We are here to help! About the root question, both you and the book are right in a way. Your book gives the positive number because, by convention, you always take the positive and real root. Thus Sqrt((-3)^2) = 3. It is important to note that while this is the primary root it is by no means the only one. In fact you can't get a positive root for problems like: ((-1)^3)^(1/3) Note that the key to this problem is raising a negative number to an odd power so it stays negative. There is clearly no positive number that you can cube to get a negative number. I mention this as a warning on the positive solution because your book is careful to pick even powers so they do not have to deal with this problem. I would also like to commend you on noticing that the square root operation yields two roots. You might be interested to know that if we consider complex numbers (ordered pairs of real numbers, which can be pictured as a plane with the real numbers on the x axis and the imaginary numbers, from sqrt(-1), on the y axis), and you take the nth root (ie x^(1/n)), you will actually get n answers! So taking cube roots in the complex plane yields 3 distinct answers, each of which when cubed yields the original number. It provides a beautiful symmetry, and is extremely useful in many applications (I am an engineer and I use this literally all the time). I find this really neat, but if you have not seen complex numbers before, fear not, you will see them all in good time. Something to look forward too. To sum up, the bottom line is your book has probably used the definition that the positive real root is the answer (this is sometimes called the primary root) when it exists. For instance, the square root of a positive number is a positive number (this is always possible). Note that the square root of a negative number would not have an answer using this definition, and it is for this case that the complex numbers were created. So your negative answers are true, they are just not what your book wanted. If you would like to see a physical interpretation of your answers do the following: 1) for Sqrt((-6)^2) write: x = Sqrt((-6)^2) 2) square both sides: x^2 = (-6)^2 3) then we have: x^2 - (-6)^2 = 0 4) graph y = x^2 - 36 and see where it crosses the x-axis. You will see it crosses at both 6 and -6. The graphing trick works because when the graph of y = x^2 - 36 crosses the x-axis we have that y = 0 so we have 0 = x^2 -36. This will reinforce the idea that both roots are valid. The reason we sometimes ignore the negative root is due to the system we are working with. Frequently we only need the positive root (in many systems we must have positive numbers such as when calculating the thickness of some material we need, since thickness measures of a part must be positive). We sometimes call these system requirements by the name constraints. One other reason that your book deals with positive roots is the simple fact that they are nicer to work with. For instance, if we consider a fourth root as a square root of a square root, we have to deal with square roots of negative numbers, which will involve complex numbers (sorry, no way around it). Hope I didn't give you too much. If something is confusing reply back as to the problem and I will try to give you some more information or a book reference if you prefer. Again, you should be very proud of your observation. Hope this helps. Good luck. -Doctor Keith, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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