# principal square root and solutions to an equation

The square root or $n^\text{th}$ root (where $n$ is even) of a number can either be $+a$ or $-a$. But for square root for example, to be a function we must define its output either to be positive or negative number because if we can get multiple values from it then it is not a function, right? So, we define the result from taking a square root for example, of a number, to be always a non-negative number. But suppose we have an equation $b^2 - c = 0$. Then, the solutions to this equation are $+\sqrt{c}$ and $-\sqrt{c}$. This means that there are two possible answers for the equation, but we want the function to have a single answer, so we agree that it will be the non-negative one. So, we first get the value from the function which is non-negative number and then we seek for the possible answers for the equation which can be positive or negative. Am i right?

## 1 Answer

I would prefer to write some of this as a comment but I'm not allowed to yet so I'll try to answer as best I can.

So you're talking about an even nth-root function, say $f(x) = \sqrt{x}$, right? Yes it's true that the principal square root is defined as the nonnegative value. So for instance $f(4) = \sqrt{4} = +2$.

I'm not really sure what it is your asking in the last part though. "So, we first get the value from the function which is non-negative number and then we seek for the possible answers for the equation which can be positive or negative. Am i right?"

This part of your question doesn't make sense without some context. Are you asking about finding the image (range) of the function? Or are you asking a question about solving the equation? If you had a specific problem in mind that might clear things up a bit.

I guess I'll mention that if you wanted to plot the negative values, then you would use the following function: $$g(x) = -\sqrt{x}$$

(notice the "$-$" in front). Then g would define the negative square roots.

If you're talking about polynomials (equations involving a variable with positive integer exponents), then to find the solutions you would first try to factor everything. In your example $b^2-c = (b-\sqrt{c})(b+\sqrt{c})=0$ which is how you can see that the two roots are $\pm \sqrt{c}$. Since you're question seems to be regarding even-powered binomials then you would always be able to factor it as a difference of squares: $(b^n-c) = (b^{n/2} - \sqrt{c})(b^{n/2} + \sqrt{c})=0$ which would usually be able to be factored further depending on your value for n.

I can explain more if you can clarify what it is you are asking.

• i am talking about solving the equation. My idea is that we define a root function to always give the positive result and when we obtain the result from the function to precede it by a minus and a plus sign because the equation has two solutions. – LearningMath Dec 9 '13 at 16:54
• I guess you could do it that way (as long as you're only using real numbers and not complex). The only reason I would think to define a function in the first place would be to plug it into a calculator or computer though. So you could use your nth-root button to find the positive answer and then tack on a negative as well if the n is even. But just be aware: if you needed the factorization of the binomial for any powers greater than 2, you wouldn't be able to find it with the function alone. – andraiamatrix Dec 9 '13 at 17:38
• I consider taking a root of a number as a function. So you have a number and when you take the root of it you get a new number. I think of this as a function, the first number as input and the new number as the output of the function. So, for taking the root of a number to be a function we must define its output. The output can be a positive or a negative number. We define the root function to output a positive number so we are done with that. Next, to get the solutions of the equation above we "call" the root function and we precede it with + and - sign denoting the two solutions. – LearningMath Dec 9 '13 at 22:09