1. ## Square Root = Two Answers

I know that there are two answers when taking the square root of a given number. For example, sqrt{4} is -2 and 2. Why is the answer both positive and negative 2? Also, does the same rule apply to variables? For example, sqrt{x^2} is x NOT
-x and x, right? Please, explain.

2. ## Re: Square Root = Two Answers

You are right about some things and wrong about others. Given any positive number x, there exist two real numbers whose square is x, one positive and one negative. We write those as $\displaystyle \pm\sqrt{x}$. But the reason we need that "$\displaystyle \pm$" is that the symbol, "$\displaystyle \sqrt{x}$" means only the positive one. That is, the square root of 4 is not "-2 and 2". The square root of 4 is 2. The two values of x, such that $\displaystyle x^2= 4$, are -2 and 2. Those are different statements. Given a positive number a, The square root of a is $\displaystyle \sqrt{a}$, a positive number. The two values of x, such that $\displaystyle x^2= a$ are $\displaystyle \sqrt{a}$ and $\displaystyle -\sqrt{x}$, which we can write as $\displaystyle \pm\sqrt{a}$. Again, the reason we need that "$\displaystyle \pm$" is that $\displaystyle \sqrt{a}$ alone refers only to the positive root of the equation.

3. ## Re: Square Root = Two Answers

Originally Posted by HallsofIvy
You are right about some things and wrong about others. Given any positive number x, there exist two real numbers whose square is x, one positive and one negative. We write those as $\displaystyle \pm\sqrt{x}$. But the reason we need that "$\displaystyle \pm$" is that the symbol, "$\displaystyle \sqrt{x}$" means only the positive one. That is, the square root of 4 is not "-2 and 2". The square root of 4 is 2. The two values of x, such that $\displaystyle x^2= 4$, are -2 and 2. Those are different statements. Given a positive number a, The square root of a is $\displaystyle \sqrt{a}$, a positive number. The two values of x, such that $\displaystyle x^2= a$ are $\displaystyle \sqrt{a}$ and $\displaystyle -\sqrt{x}$, which we can write as $\displaystyle \pm\sqrt{a}$. Again, the reason we need that "$\displaystyle \pm$" is that $\displaystyle \sqrt{a}$ alone refers only to the positive root of the equation.
Ok. Now, the sqrt{-4} = -2i and 2i.
The solution set is {-2i, 2i}.
Why does the sqrt{-1} = i? Why does i^2 = -1?

4. ## Re: Square Root = Two Answers

Originally Posted by harpazo
Why does the sqrt{-1} = i? Why does i^2 = -1?
The imaginary unit "i" is defined as $\displaystyle i^2 = -1$ so there is nothing to derive.

-Dan

5. ## Re: Square Root = Two Answers

Originally Posted by harpazo
Ok. Now, the sqrt{-4} = -2i and 2i.
The solution set is {-2i, 2i}.
Why does the sqrt{-1} = i? Why does i^2 = -1?
@harpazo, if you understand the use of definitions & axioms in mathematics this should help.
The equation $x^2+1=0$ has no solutions in the system of real numbers because $x^2\ge 0$ for all real numbers $x$.
So we postulate the existence, the addition, of one new number $\bf\mathit{i}$ the is a solution for $x^2+1=0$.
That leads to the conclusion that $\bf\mathit{i~}^2=-1$. Moreover also that $-\bf\mathit{i}$ is also a solution for $x^2+1=0$

6. ## Re: Square Root = Two Answers

Originally Posted by Plato
@harpazo, if you understand the use of definitions & axioms in mathematics this should help.
The equation $x^2+1=0$ has no solutions in the system of real numbers because $x^2\ge 0$ for all real numbers $x$.
So we postulate the existence, the addition, of one new number $\bf\mathit{i}$ the is a solution for $x^2+1=0$.
That leads to the conclusion that $\bf\mathit{i~}^2=-1$. Moreover also that $-\bf\mathit{i}$ is also a solution for $x^2+1=0$
Good information for my notes.