1.math work!!!!!!!!!! When do you get "does not exist" using square roots. I know -x^(1/2) returns an does not exist. (Integer x>0) Is there any more possible undefines for roots? Is -9^(1/4) an error? ----------------------------------------------------------------- 2. How do you do this? 90 decrease by 7%.

You cannot take the square root of a negative. Those are imaginary numbers. -x^(1/n) where n is any even number will return an error on your calculator. 90*(1 - Percent/100) 90*(1 - 0.07) = 83.7

Ok, here's the deal on roots of numbers. First of all, the example you gave, exactly as it's written, -x^(1/2) is NOT undefined. This is because of the order of operations. An expansion to the formula given is (-1) * x^(1/2). You multiply by the negative at the very end, so this shouldn't be a problem given that x is greater than zero (or even equal to zero). However, what you meant must've been (-x)^(1/2), which, given x is positive will be undefined. One thing I should note here is that (-x)^(1/2) is only undefined when x is strictly positive, meaning zero's not included. This is for obvious reasons, -0 = 0, and 0^(1/2) is defined. So, why is this? Why can't we take the root of a negative number? Well, let's look at the problem another way. When you're asking for the square root of a number, say, x, you're asking yourself, "What number multiplied by itself gives me x?" Now, there are only three cases for this question. First, let's take care of the easiest case, where the root is 0. What happens here should be very obvious. The next case is if the the root is positive. If we multiply a positive number by another positive number (even if they're not the same, keeping in mind zero is not considered positive or negative) we'll always get a positive number back. No matter what. The third case is when the root is negative. A negative number, multiplied by any other negative number, is POSITIVE! Well, what does this tell us? It really just says "Any number, multiplied by itself, will never give us a negative value". So, going back to the original question, "What number multiplied by itself gives me x," if x is negative we can easily say that there's no such number that gives us x when it's multiplied by itself. More formally, in math, when this happens, we say the result is "undefined". So what happens when we DO want to take the square root of a negative number? Well, first let's examine a property about square roots. sqrt(a*b) = sqrt(a) * sqrt(b). This is true all the time. When we look at a negative number, we can really look at it as sqrt(-1 * x), where x is positive. sqrt(-1 * x) = sqrt(-1) * sqrt(x), and we CAN figure out sqrt(x). So the part that gives us problems is the sqrt(-1). This being the case, some guy in history just said "Hey, you know what, let's just make up a new number that is defined as sqrt(-1), then we can do roots of negative numbers!" That number is given the symbol i. Or sometimes j, depending on why you're using roots of negative numbers. But it's usually i, from what I've seen. So then sqrt(-x) = i*sqrt(x). This next paragraph's just a little about complex numbers and their relationship with real numbers, feel free to skip ahead if you don't really care. Any number that uses that i that we were talking about is called a complex number. At the same time, this also means that any number that we can write using i can also be considered a complex number. Complex numbers are of the form a + bi, where a and b are real numbers. If you hadn't already guessed, real numbers ARE complex numbers too! This is because real numbers can be written in a + bi form, as long as b = 0. If this is a little hard to understand, just consider an integer. An integer is, of course, an integer. But every integer is also a rational number, because it can be written as a fraction, right? And all rational numbers are real numbers, so really, any integer is an integer, a rational, and a real number at the same time. We can do all sorts of cool stuff with complex numbers, but that's for another time.

sqrt(any negative #) is an imaginary number. but if the deal right now is sqrt(-x) the only number x can take for your calculator to work is if the x is a negative number or 0. So in general, the domain of the function f(x) = sqrt(-x) is from -infinity to 0. 4th root(-9) is also an imaginary number so you'll get an error. But the 3rd root (-9) will work. For you I guess it'll look like this (-9)^(1/3)