# surd algebra help

• May 29th 2009, 08:17 AM
Tweety
surd algebra help
express $\displaystyle \sqrt{22.5}$ in the form $\displaystyle K\sqrt{10}$

I dont get how to do this, can some one show me? Thanks.
• May 29th 2009, 08:25 AM
e^(i*pi)
Quote:

Originally Posted by Tweety
express $\displaystyle \sqrt{22.5}$ in the form $\displaystyle K\sqrt{10}$

I dont get how to do this, can some one show me? Thanks.

Recall that
$\displaystyle \sqrt{ab} = \sqrt{a} \sqrt{b}$

$\displaystyle \sqrt{a^2} = a$

$\displaystyle \sqrt{22.5} = K\sqrt{10} = \sqrt{10K^2}$

Square both sides:

$\displaystyle 10K^2 = 22.5$

solve for K
• May 29th 2009, 08:26 AM
$\displaystyle \sqrt{22.5} = \sqrt{2.25 \cdot 10} = \sqrt{2.25} \cdot \sqrt{10}$
So $\displaystyle K=\sqrt{2.25} = \frac{3}{2}$
• May 29th 2009, 08:30 AM
masters
Quote:

Originally Posted by Tweety
express $\displaystyle \sqrt{22.5}$ in the form $\displaystyle K\sqrt{10}$

I dont get how to do this, can some one show me? Thanks.

Hi Tweety,

Another approach:

$\displaystyle \sqrt{22.5}=\sqrt{\frac{225}{10}}=15\sqrt{\frac{1} {10}}=\frac{15\sqrt{10}}{10}=\frac{3}{2}\sqrt{10}$
• May 29th 2009, 08:30 AM
Tweety
the correct answer is $\displaystyle \frac{3}{2}\sqrt{10}$
edit; oh right I get it now, thanks!
• May 29th 2009, 08:35 AM
e^(i*pi)
Quote:

Originally Posted by Tweety
the correct answer is $\displaystyle \frac{3}{2}\sqrt{10}$
edit; oh right I get it now, thanks!

The square root of 2.25 is 1.5 (Wink)