# surds

• Jun 4th 2009, 12:05 AM
b3nj1d
surds
I am having difficulty solving these surds problems

$\displaystyle (5+2 \sqrt{7})^2$

and

$\displaystyle \sqrt{8}+\sqrt{18}-\sqrt{50}$
• Jun 4th 2009, 12:24 AM
alexmahone
$\displaystyle (5+2\sqrt7)^2=25+28+20\sqrt{7}$
$\displaystyle =53+20\sqrt{7}$

$\displaystyle \sqrt{8}+\sqrt{18}-\sqrt{50}=2\sqrt{2}+3\sqrt{2}-5\sqrt{2}=0$
• Jun 4th 2009, 12:30 AM
Prove It
Quote:

Originally Posted by b3nj1d
I am having difficulty solving these surds problems

$\displaystyle (5+2 \sqrt{7})^2$

and

$\displaystyle \sqrt{8}+\sqrt{18}-\sqrt{50}$

Since we haven't learnt FOIL yet, I understand why you're having trouble with the first one.

If you have two binomials like $\displaystyle (a + b)(c + d)$, they are expanded by multiplying the Firsts, then the Outers, then the Inners, then the Lasts.

So $\displaystyle (a + b)(c + d) = ac + ad + bc + bd$.

So for your first one, you have

$\displaystyle (5 + 2\sqrt{7})^2 = (5 + 2\sqrt{7})(5 + 2\sqrt{7})$

Multiplying the firsts gives $\displaystyle 5 \times 5 = 25$.

Multiplying the outers gives $\displaystyle 5 \times 2\sqrt{7} = 10\sqrt{7}$

Multiplying the inners gives $\displaystyle 5 \times 2\sqrt{7} = 10\sqrt{7}$

Multiplying the lasts gives $\displaystyle 2\sqrt{7}\times 2\sqrt{7} = 4 \times 7 = 28$.

So $\displaystyle (5 + 2\sqrt{7})^2 = 25 + 10\sqrt{7} + 10\sqrt{7} + 28$

$\displaystyle = 53 + 20\sqrt{7}$.

For the second, you need to simplify all of the surds first. Then hopefully you'll have "like surds" that can be collected.

$\displaystyle \sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}$.

$\displaystyle \sqrt{18} = \sqrt{9 \times 2} = \sqrt{9}\times\sqrt{2} = 3\sqrt{2}$.

$\displaystyle \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25}\times\sqrt{2} = 5\sqrt{2}$.

So $\displaystyle \sqrt{8} + \sqrt{18} - \sqrt{50} = 2\sqrt{2} + 3\sqrt{2} - 5\sqrt{2}$

$\displaystyle = 5\sqrt{2} - 5\sqrt{2}$

$\displaystyle = 0\sqrt{2}$

$\displaystyle = 0$.

Chookas for the exam tomorrow.