# Problem #2 - Rationalizing

• Apr 20th 2014, 10:49 AM
topsquark
Problem #2 - Rationalizing
This is for the Pre-Calculus students.

A common way to add fractions with radicals in the denominator is to "rationalize the denominator." For example, rationalize the denominator of
$\displaystyle \frac{2x + 3y}{\sqrt{x} - \sqrt{y} }$

To rationalize this we use $\displaystyle (a - b)(a + b) = a^2 - b^2$ and multiply the numerator and denominator by $\displaystyle \sqrt{x} + \sqrt{y}$ giving

$\displaystyle \frac{2x + 3y}{\sqrt{x} - \sqrt{y} } \cdot \frac{\sqrt{x} + \sqrt{y}}{\sqrt{x} + \sqrt{y}} = \frac{(2x + 3y) \cdot ( \sqrt{x} + \sqrt{y} ) }{x - y}$

How do you rationalize the denominator of this?
$\displaystyle \frac{2x + 3y}{\sqrt[3]{x} + \sqrt[3]{y}}$

Hint: How do you factor $\displaystyle a^3 + b^3$?

-Dan
• Apr 20th 2014, 05:36 PM
Soroban
re: Problem #2 - Rationalizing
Hello, topsquark!

Quote:

How do you rationalize the denominator of this?
$\displaystyle \frac{2x + 3y}{\sqrt[3]{x} + \sqrt[3]{y}}$

Hint: How do you factor $\displaystyle a^3 + b^3$?

Spoiler:

We have: .$\displaystyle \frac{2x+3y}{x^{\frac{1}{3}} + y^{\frac{1}{3}}}$

Multiply by $\displaystyle \frac{x^{\frac{2}{3}} - x^{\frac{1}{3}}y^{\frac{1}{3}} + y^{\frac{2}{3}}}{x^{\frac{2}{3}} - x^{\frac{1}{3}}y^{\frac{1}{3}} + y^{\frac{2}{3}}}$

. . $\displaystyle \frac{2x+3y}{x^{\frac{1}{3}} + y^{\frac{1}{3}} }\cdot\frac{x^{\frac{2}{3}} - x^{\frac{1}{3}}y^{\frac{1}{3}} + y^{\frac{2}{3}}}{x^{\frac{2}{3}} - x^{\frac{1}{3}}y^{\frac{1}{3}} + y^{\frac{2}{3}}} \;=\; \frac{(2x+3y)(x^{\frac{2}{3}} - x^{\frac{1}{3}}y^{\frac{1}{3}} + y^{\frac{2}{3}})}{x+y}$

• Apr 27th 2014, 04:00 AM
topsquark
re: Problem #2 - Rationalizing
And a big "nicely done" to Soroban for solving this week's challenge. His prize for submitting a correct entry is new car! (If I can still find those chocolate cars at Walgreen's.)

-Dan