# Rationals

• Nov 6th 2007, 03:05 PM
nathan02079
Rationals

How do i simplify...

$\displaystyle 2\sqrt 150 \over \sqrt 8$

Simplify by rationalizing the denominator

$\displaystyle \sqrt 32 \over \sqrt 18$

Express the following with rational denominators

$\displaystyle \sqrt 7 - \sqrt 2 \over \sqrt 2$

Simplify and express in lowest terms

$\displaystyle 10 \sqrt 18 - 5 \sqrt 24 \over \sqrt 5$

any help would be appreciated! THANK YOU!
• Nov 6th 2007, 03:32 PM
Plato
$\displaystyle \begin{array}{rcl} \frac{{10\sqrt {18} - 5\sqrt {24} }}{{\sqrt 5 }} & = & \frac{{10\sqrt {2 \cdot 9} - 5\sqrt {4 \cdot 6} }}{{\sqrt 5 }} \\ & = & \frac{{30\sqrt 2 - 10\sqrt 6 }}{{\sqrt 5 }} \\ & = & \frac{{30\sqrt {10} - 10\sqrt {30} }}{5} \\ & = & 6\sqrt {10} - 2\sqrt {30} \\ \end{array}$
• Nov 6th 2007, 04:13 PM
Soroban
Hello, nathan02079!

Quote:

$\displaystyle 2\sqrt 150 \over \sqrt 8$
We have: .$\displaystyle 2\sqrt{\frac{150}{8}} \;=\;2\sqrt{\frac{75}{4}} \;=\;2\frac{\sqrt{75}}{2} \;=\;\sqrt{75} \;=\;\sqrt{25\cdot3} \;=\;\sqrt{25}\cdot\sqrt{3} \;=\;5\sqrt{3}$

Quote:

Simplify by rationalizing the denominator:

$\displaystyle \sqrt 32 \over \sqrt 18$

We have: .$\displaystyle \frac{\sqrt{32}}{\sqrt{18}} \;=\;\sqrt{\frac{32}{18}} \;=\;\sqrt{\frac{16}{9}} \;=\;\frac{\sqrt{16}}{\sqrt{9}} \;=\;\frac{4}{3}$

Quote:

Express the following with rational denominators

$\displaystyle \sqrt 7 - \sqrt 2 \over \sqrt 2$

Rationalize: .$\displaystyle {\color{blue}\frac{\sqrt{2}}{\sqrt{2}}}\cdot\frac{ \sqrt{7} - \sqrt{2}}{\sqrt{2}} \;=\;\frac{\sqrt{14} - 2}{2}$