# Math Help - surds

1. ## surds

I am having difficulty solving these surds problems

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

and

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

2. $(5+2\sqrt7)^2=25+28+20\sqrt{7}$
$=53+20\sqrt{7}$

$\sqrt{8}+\sqrt{18}-\sqrt{50}=2\sqrt{2}+3\sqrt{2}-5\sqrt{2}=0$

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

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

and

$\sqrt{8}+\sqrt{18}-\sqrt{50}$
Since we haven't learnt FOIL yet, I understand why you're having trouble with the first one.

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

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

So for your first one, you have

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

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

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

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

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

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

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

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

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

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

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

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

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

$= 0\sqrt{2}$

$= 0$.

Chookas for the exam tomorrow.