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:


The answer is:


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


But I end up with: $$\frac{\sqrt[7]{x^4}}{x^4}$$

What am I doing wrong?

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

You have to multiply it by $\frac {\sqrt [7]{x^3}}{\sqrt [7]{x^3}}$ The numerator in the answer shows that. Then ${\sqrt [7]{x^3}}\cdot {\sqrt [7]{x^4}}={\sqrt [7]{x^7}}=1$. I find it easier to use fractional exponents, changing ${\sqrt [7]{x^4}}$ to $x^{\frac 47}$ and proceeding. You could do that and see if it helps.

  • $\begingroup$ It's coming back to me now. I need to fill in the fraction basically. $\endgroup$ – munchschair Dec 7 '13 at 1:10

What you're doing wrong is that you're following a rule that says you should multiply the numerator and denominator by $\sqrt[7]{x}$.

One the bottom you've got $$ \sqrt[7]{x}\cdot\sqrt[7]{x}\cdot\sqrt[7]{x}\cdot\sqrt[7]{x}. $$ What you need on the bottom to get rid of the radical is $$ \sqrt[7]{x}\cdot\sqrt[7]{x}\cdot\sqrt[7]{x}\cdot\sqrt[7]{x}\cdot\sqrt[7]{x}\cdot\sqrt[7]{x}\cdot\sqrt[7]{x}. $$ Multiply that out an you just have $x$, with no radical.

So you have to multiply both the numerator and the denominator by $\sqrt[7]{x}^3$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.