Rewrite radicals and rational exponents

Start with something you know: $\sqrt9=3$. In exponent form, that's $9^{1/2}=3$. A power of $\frac12$ means a square root, and a power of $\frac13$ means a cube root, so $8^{1/3}=\sqrt[3]{8}=2$. An exponent that's a fraction like this is called a rational exponent.

When the top of the fraction isn't $1$, it adds a power. Take $8^{2/3}$. The $3$ says take the cube root, and the $2$ says square the result: $\left(\sqrt[3]{8}\right)^2=2^2=4$. You can square first and get the same answer: $\sqrt[3]{8^2}=\sqrt[3]{64}=4$.

Here's the same idea written for any base. For $x>0$ and a whole number $n>1$,

And when the top is any integer $m$,

The bottom number, $n$, tells you the root. The top number, $m$, tells you the power. A short way to hold onto it is power over root:

Every question here says $x>0$, and there's a reason. Even roots can hide a sign. If $x$ could be negative, then $\sqrt{x^2}=|x|$. For example, $\sqrt{(-3)^2}=\sqrt9=3$, not $-3$. When $x>0$, $\sqrt{x^2}=x$, so you can focus on the exponents without an absolute-value case.

Try it with numbers first. $(4\cdot9)^{1/2}=\sqrt{36}=6$. Now give each factor its own square root: $4^{1/2}\cdot9^{1/2}=2\cdot3=6$. Same answer. That's the rule: when a rational exponent sits outside a product, it applies to every factor in that product. For positive values,

For example, the exponent $\frac13$ reaches the $27$, the $a^6$ and the $b^3$:

A negative outside exponent works the same way. Apply it to every factor first. Then any factor that ends up with a negative exponent becomes a reciprocal, so $y^{-4}$ turns into $\frac{1}{y^4}$.

One warning: this works for products, not sums. $(a+b)^r$ usually isn't $a^r+b^r$. A quick number check shows it: $(9+16)^{1/2}=\sqrt{25}=5$, but $9^{1/2}+16^{1/2}=3+4=7$.

To simplify a radical, split what's inside into two parts: a perfect-power factor, which the root can undo completely, and whatever's left over. Under a square root, look for a perfect square. Under a cube root, look for a perfect cube, and so on for higher roots.

For $x>0$,

Since $36x^4=(6x^2)^2$, it comes out of the radical as $6x^2$. The $2x$ isn't a perfect square, so it stays inside.

The same idea works in exponent form. Take $\sqrt{x^5}=x^{5/2}$. Since $\frac52=2+\frac12$, you get $x^{5/2}=x^2\cdot x^{1/2}=x^2\sqrt{x}$. The whole-number part of the exponent comes out, and the fraction stays under the root.

Reach for full-curve overlap first when the question has one variable, every choice is a finished expression you can type in, and there are enough products or roots that rewriting by hand would take a while. Graph only the $x$-values the question allows, like $x>0$. Then make sure the curves match along a whole stretch of the graph, not just meet at one point.

Work by hand instead in these cases:

• When there's more than one variable, like the $x$ and $y$ in the mixed example, or a constant that could be any value, rewrite with exponents.
• When one short extraction settles it, like $\sqrt{50x^3}=\sqrt{25x^2\cdot2x}=5x\sqrt{2x}$, doing it by hand is faster than typing it in.
• When the question asks you to write an exact form yourself, or to explain why a form is right, the graph can't write it for you.
• When the answer has to include a restriction, or a value $x$ can't take, you need algebra to find it.

Why does a whole curve count when one point doesn't? Plugging in a single number can rule a choice out, but two different expressions can still agree at that number. For example, $x^2$ and $2x$ both equal $4$ at $x=2$. Full-curve overlap compares the two expressions at every allowed $x$ on the graph, not at one, so it's much stronger evidence.