surds

**b3nj1d** · June 4th 2009, 12:05 AM

I am having difficulty solving these surds problems

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

and

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

**alexmahone** · June 4th 2009, 12:24 AM

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

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

**Prove It** · June 4th 2009, 12:30 AM

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.

Thread: surds

LinkBack

Thread Tools

Display

surds

Similar Math Help Forum Discussions

[SOLVED] Help with Surds

Something to do with surds?

surds

surds

Surds

Search Tags