1. ## Surds

I understand the basic idea of getting the sq.rt off the bottom by multiplying top and bottom by the sq.rt but it doesn't always seemto work, is there more to it?

2/sq.rt8 -I get sq.rt8/4 , but answer says sq.rt2/2

a/ sq.rt40/2 -I get sq.rt4/8 -answer says sq.rt10a/10

And I am not sure how to tackle this one:

2/ 1+sq.rt6

Thanks.

I understand the basic idea of getting the sq.rt off the bottom by multiplying top and bottom by the sq.rt but it doesn't always seemto work, is there more to it?

2/sq.rt8 -I get sq.rt8/4 , but answer says sq.rt2/2

a/ sq.rt40/2 -I get sq.rt4/8 -answer says sq.rt10a/10

And I am not sure how to tackle this one:

2/ 1+sq.rt6

Thanks.
For the first one:
$\frac{2}{\sqrt{8}} \cdot \frac{\sqrt{8}}{\sqrt{8}} = \frac{2 \sqrt{8}}{8}$

$= \frac{\sqrt{8}}{4}$

So far, so good. But:
$\frac{ \sqrt{8}}{4} = \frac{ \sqrt{4 \cdot 2}}{4} = \frac{2 \sqrt{2}}{4} = \frac{ \sqrt{2}}{2}$

For the second one: " a/ sq.rt40/2"
$\frac{a}{\frac{\sqrt{40}}{2}} = \frac{2a}{\sqrt{40}} = \frac{2a}{\sqrt{40}} \cdot \frac{\sqrt{40}}{\sqrt{40}}$

$= \frac{2a\sqrt{40}}{40} = \frac{a\sqrt{40}}{20}$

$= \frac{a \sqrt{4 \cdot 10}}{20} = \frac{2a\sqrt{10}}{20} = \frac{a\sqrt{10}}{10}$

For the last one: "2/ 1+sq.rt6"
$\frac{2}{1 + \sqrt{6}}$

We need to multiply the top and bottom by the "conjugate." (The works on the ones above too.) It works like this:
$(a + b)(a - b) = a^2 - b^2$

So
$(a + \sqrt{c})(a - \sqrt{c}) = a^2 - c$
which no has no square root in the final expression.

Thus if we have a denominator with a $a \pm \sqrt{c}$ in it we multiply the top and bottom by $a \mp \sqrt{c}$, which is called the conjugate of $a \pm \sqrt{c}$.

So.
$\frac{2}{1 + \sqrt{6}} \cdot \frac{1 - \sqrt{6}}{1 - \sqrt{6}} = \frac{2 - 2 \sqrt{6}}{1^2 - 6} = \frac{2 - 2\sqrt{6}}{-5}$

$= - \frac{2 - 2\sqrt{6}}{5} = \frac{-2 + 2\sqrt{6}}{5}$

-Dan

$\frac{2}{\sqrt{8}}$ . . . I get $\frac{\sqrt{8}}{4}$, but answer says $\frac{\sqrt{2}}{2}$ . . . . Reduce your answer

Here's something I figured out when I was a student
. . and I've taught it to all my students ever since.
Yet I have never seen it mentioned in any textbook
. . nor have I ever met a teacher who is familiar with it.

The denominator has a square root so we're expected to "rationalize" it.
. . Very well, how do we do this?

We modify the fraction (legally) so that the square root contains a perfect square.

The standard method is to multiply top and bottom by the square root:

. . $\frac{2}{\sqrt{8}}\cdot\frac{\sqrt{8}}{\sqrt{8}} \;=\;\frac{2\sqrt{8}}{\underbrace{\sqrt{64}}_{\tex t{square}}} \;=\;\frac{2\sqrt{8}}{8}$

It is rationalized, but we are (of course) expected to reduce the fraction.
. . Obviously, we can "cancel 2's" and get $\frac{\sqrt{8}}{4}$

But $\sqrt{8}$ can be reduced. .(We're supposed to be aware of this.)

Since $\sqrt{8} \:=\:\sqrt{4\cdot2} \:=\:\sqrt{4}\!\cdot\!\sqrt{2}\:=\:2\sqrt{2}$, we have: . $\frac{\sqrt{8}}{4}\:=\:\frac{2\sqrt{2}}{4} \:=\:\frac{\sqrt{2}}{2}$

But wouldn't be simplier to multiply top and bottom by $\sqrt{2}$ ?

. . $\frac{2}{\sqrt{8}}\cdot\frac{\sqrt{2}}{\sqrt{2}} \;=\;\frac{2\sqrt{2}}{\underbrace{\sqrt{16}}_{\tex t{square}}} \;=\;\frac{2\sqrt{2}}{4} \:=\:\frac{\sqrt{2}}{2}$

My point: Multiply top and bottom by the smallest amount
. . . . . . . to make the denominator a square.

Example: . $\frac{1}{\sqrt{12}}$

Instead of automatically multiplying top and bottom by $\sqrt{12}$,
. . pause and give it some thought.

If we introduce a "3", we will have a square: . $\frac{1}{\sqrt{12}}\cdot\frac{\sqrt{3}}{\sqrt{3}} \:=\:\frac{\sqrt{3}}{\sqrt{36}} \:=\:\frac{\sqrt{3}}{6}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This problem arises because the square in the denominator can be simplified.

An alternate approach is to simplify the denominator first:

. . $\frac{1}{\sqrt{12}} \;=\;\frac{1}{2\sqrt{3}}$ . . . then rationalize: . $\frac{1}{2\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} \;=\;\frac{\sqrt{3}}{2\sqrt{9}}\;=\;\frac{\sqrt{3} }{6}$

4. Thanks for all the previous replies, very very useful.

I had a go at: 6/ 3 + sq.rt3

= 6.(3-sq.rt3 / (3+sq.rt3)(3-sq.rt3)

= 18 - 6sq.rt3 / 9 - 3

= 3-6sq.rt3

However, the answer says 3 -sq.rt3 ??

Another one:

sq.rt18 - sq.rt9

= (sq.rt9 * sq.rt2) -3

= 3sq.rt2 - 3 -is this right?

I had a go at: . $\frac{6}{3 + \sqrt{3}}$

$= \;\frac{6(3 -\sqrt{3})}{(3 + \sqrt{3})(3 - \sqrt{3})}$

$= \;\frac{18 - 6\sqrt{3}}{9 - 3}$ . . . . Why distribute the 6?

$= \;3 - 6\sqrt{3}$ . . . . How?

Your third line would have been: . $\frac{6(3-\sqrt{3})}{9-3} \;=\;\frac{\not{6}(3-\sqrt{3})}{\not{6}} \;=\;3-\sqrt{3}$

Another one: . $\sqrt{18} - \sqrt{9} \;=\;\sqrt{9}\cdot\sqrt{2} - 3 \;=\;3\sqrt{2} - 3$

Is this right?
. . . Yes!

I had a go at: 6/ 3 + sq.rt3
[/quote]
Just a quick note. Please try to use parenthesis. What you intended was
$\frac{6}{3 + \sqrt{3}}$

but according to order of operations, division comes before addition, so what you wrote was
$\frac{6}{3} + \sqrt{3}$

This could be the source of some confusion later on. If you don't learn the LaTeX then you should write it like this: 6/(3 + sq.rt3)

-Dan