Some powers \(a^b\) have common names. A square is an expression of the form \(a^2\text{,}\) because this is the area of a square with side length \(a\text{.}\) A cube is an expression of the form \(a^3\text{,}\) because this is the volume of a cube with side length \(a\text{.}\)
We call \(3\) and \(-3\) square roots of \(9\text{.}\) Notice I said “square roots” and not “the square root”! One of these, \(3\text{,}\) is positive; and the other, \(-3\text{,}\) is negative. We use the symbol \(\sqrt{9}\) specifically for the positive square root and call it the principal square root of \(9\text{.}\) So \(\sqrt{9} = 3\text{.}\)
Note that the other square root of \(9\) is the opposite of its principal square root, namely \(-\sqrt{9} = -3\text{.}\) There are exactly two square roots of \(9\) and they are \(\sqrt{9}\) and \(-\sqrt{9}\text{.}\) This is summarized by saying the square roots of \(9\) are \(\pm\sqrt{9}\text{,}\) where the \(\pm\) is read as “plus or minus”.
Now consider \(-9\text{,}\) a negative number. Since the square of any real number is never negative, \(-9\) has no real number square roots and the symbol \(\sqrt{-9}\) is undefined. To put things another way, the equation \(x^2 = -9\) has no real number solutions. 1
Well, you may have seen complex numbers which will give meaning to such an expression, but we won’t discuss that here.
Finally, observe that \(0^2 = 0\text{,}\) so that \(0\) is a square root of itself. This is the only square root of zero, so that \(\sqrt{0} = 0\text{.}\)
The equation \(x^2 = 15\) has exactly two real number solution \(x=\pm \sqrt{15}\text{.}\) These numbers are irrational, but may be approximated with a calculator or computer. Using the square root button, I found that \(\sqrt{15} \approx 3.87\text{,}\) approximately. The solutions are approximately \(x \approx 3.87\) and \(x \approx - 3.87\text{.}\) That is,
Be careful simplifying the expression \(\sqrt{x^2}\text{.}\) First, note that \(x\) is squared, making the resulting \(x^2\) non-negative. Then we want the principal square root of \(x^2\text{.}\) Consider two examples, first with \(x=4\)
which does not result in \(-4\text{.}\) In both cases, we finished with the positive value \(4\text{,}\) i.e. the absolute value of \(x\text{.}\) In general, for any real number \(x\text{,}\)
Try to avoid viewing \(\sqrt{\quad}\) as an operation you perform to both sides. Often students say “take the square root of both sides” but usually end up only writing the principal square root. Doing this leads to missing the other square root of \(15\text{,}\) namely \(-\sqrt{15}\text{.}\) Instead, think of it as an if/then sentence: "if \(x^2 = 15\text{,}\) then \(x = \pm \sqrt{15}\) ".
The number \(-8\) is the cube of \(-2\text{,}\) that is, \((-2)^3 = 8\text{.}\) This is the only real number with this property, so we write \(\sqrt[3]{-8} = -2\) for the principal cube root. In general, every real number, including negative numbers, has exactly one real number cube root. To put things another way, there is exactly one real number solutions to the equation \(x^3 = -8\text{,}\) namely \(x = -2\text{.}\)
The expression \(\sqrt{x}\) denotes the principal square root of \(x\text{.}\) If \(x \geq 0\text{,}\) then \(\sqrt{x} \geq 0\) and satisfies \((\sqrt{x})^2 = x\text{.}\) If \(x \lt 0\text{,}\) then \(\sqrt{x}\) is undefined.
