# Rationalizing Denominator question

So I feel totally stupid asking this considering I am in precalc but our professor threw this question at us. I have the answer, but he didn't provide the process:

Rationalize the denominator:

$$\frac{1}{\sqrt{x^4}}$$

$$\frac{\sqrt{x^3}}{x}$$

Please explain it how he got there. I know you have to multiply by

$$\frac{\sqrt{x^4}}{\sqrt{x^4}}$$

But I end up with: $$\frac{\sqrt{x^4}}{x^4}$$

What am I doing wrong?

• What if you write it as: $\dfrac{x^{3/7}}{x^1} = \dfrac{x^{3/7}}{x^{7/7}}$? Does that help? – Amzoti Dec 7 '13 at 0:55

You have to multiply it by $\frac {\sqrt {x^3}}{\sqrt {x^3}}$ The numerator in the answer shows that. Then ${\sqrt {x^3}}\cdot {\sqrt {x^4}}={\sqrt {x^7}}=1$. I find it easier to use fractional exponents, changing ${\sqrt {x^4}}$ to $x^{\frac 47}$ and proceeding. You could do that and see if it helps.
What you're doing wrong is that you're following a rule that says you should multiply the numerator and denominator by $\sqrt{x}$.
One the bottom you've got $$\sqrt{x}\cdot\sqrt{x}\cdot\sqrt{x}\cdot\sqrt{x}.$$ What you need on the bottom to get rid of the radical is $$\sqrt{x}\cdot\sqrt{x}\cdot\sqrt{x}\cdot\sqrt{x}\cdot\sqrt{x}\cdot\sqrt{x}\cdot\sqrt{x}.$$ Multiply that out an you just have $x$, with no radical.
So you have to multiply both the numerator and the denominator by $\sqrt{x}^3$.