# Help with Surds

• Jan 5th 2012, 12:09 PM
ashleysmithd
Help with Surds
Hi,

I need a bit of help solving a question on surds if anyone can push me in the right direction with it.

The question is:
http://dl.dropbox.com/u/16469323/question.jpg

I've started multiplying out the bottom row, and have got as far as this:

$\frac{4\sqrt{3}+3\sqrt{7}}{3\sqrt{3}+\sqrt{7}} * \frac{3\sqrt{3}-\sqrt{7}}{3\sqrt{3}-\sqrt{7}}$

I then proceed to multiplying out the bottom row;

$3*3+3\sqrt{3}-3\sqrt{7}+3\sqrt{3}+\sqrt{3}*\sqrt{3}-\sqrt{21}+3\sqrt{7}+\sqrt{21}-7$

I then proceed to collecting the like terms;

$(9+3-7)+(-3\sqrt{7}+3\sqrt{7})+(-\sqrt{21}+\sqrt{21})+(3\sqrt{3}+3\sqrt{3})$

$= 5 + 6\sqrt{3}$

From here, I really can't see how to 'rationalise the denominator'.

I'm only working on the bottom row right now, but if anyone can tell me where I'm going wrong with this that would be great. I'm sure it's something simple, I'm just not seeing it.

• Jan 5th 2012, 12:37 PM
earboth
Re: Help with Surds
Quote:

Originally Posted by ashleysmithd
Hi,

I need a bit of help solving a question on surds if anyone can push me in the right direction with it.

The question is:
http://dl.dropbox.com/u/16469323/question.jpg

I've started multiplying out the bottom row, and have got as far as this:

$\frac{4\sqrt{3}+3\sqrt{7}}{3\sqrt{3}+\sqrt{7}} * \frac{3\sqrt{3}-\sqrt{7}}{3\sqrt{3}-\sqrt{7}}$

I then proceed to multiplying out the bottom row;

...

Not sure what you've done in the next steps.

Use (a + b)(a - b) = aČ - bČ

That means the denominator becomes:

$({3\sqrt{3}+\sqrt{7}}) \cdot ({3\sqrt{3}-\sqrt{7}})=9\cdot 3 - 7 =20$

then:
- expand the numerator
- collect those terms with $\sqrt{21}$ and those with simple integers
- factor out 5 at numerator and denominator
- cancel the common factor

You should come out with $\frac{3+\sqrt{21}}4$
• Jan 5th 2012, 01:04 PM
ashleysmithd
Re: Help with Surds
Thanks for the reply, this is starting to make sense.

The only confusion I still have is how multiplying out the numerator becomes $15+\sqrt{21}$

$(4*3)*3 = 36$
$(3\sqrt{7}*-\sqrt{7}) = -21$
$\Rightarrow 36 - 21$ to me is making 15, only.

In what order is it multiplied out in to make $15+\sqrt{21}$?

Thanks again.
• Jan 5th 2012, 01:33 PM
pickslides
Re: Help with Surds
So you need to expand this guy?

$(4\sqrt{3}+3\sqrt{7})(3\sqrt{3}-\sqrt{7})$

Follow this method $(a+b)(c+d) = ac+ad+bc+bd$

Does this make sense?
• Jan 5th 2012, 02:19 PM
ashleysmithd
Re: Help with Surds
Ah, yes. I was trying to use the $(a+b)(c+d) = ac + ad + bc + bd$ before except I was separating the coefficients from the roots, completely, hence why my multiplying out was so long because I was multiplying through several times too many. Basically, I didn't understand that $3\sqrt{7}$ was a single term, I was treating the $3$ and $\sqrt{7}$ as if they were separate.

Many thanks.

There's just one last thing I don't get... For the numerator, I got $15+5\sqrt{21}$. Where does the 5 go? I know earboth said about factoring it out at the numerator and denominator, but I'm not sure I understand.
• Jan 5th 2012, 02:33 PM
pickslides
Re: Help with Surds
$15+5\sqrt{21} = 5\times 3 +5\times \sqrt{21} = 5(3 + \sqrt{21})$ now what did you get for the denominator?
• Jan 5th 2012, 02:51 PM
ashleysmithd
Re: Help with Surds
I got $20$ for the denominator, I worked this out through;

$(3\sqrt{3}+\sqrt{7})(3\sqrt{3}-\sqrt{7})$

$= 9\sqrt{9} -3\sqrt{21}+3\sqrt{21}-7$
$= 9(3)-7$
$= 27 - 7 = 20$

I'm still a bit unsure about the last bit, where the 5 goes. I can understand that $15+5\sqrt{21}$ is equivalent to $5(3+\sqrt{21})$, but multiplied out that still gives $15+5\sqrt{21}$.

Apologies, I really don't get this last bit.
• Jan 5th 2012, 02:56 PM
pickslides
Re: Help with Surds
Ok, you have done all the hard work here, just need to cancel some terms,

$\frac{15+5\sqrt{21}}{20} = \frac{5\times 3 +5\times \sqrt{21}}{5\times 4} = \frac{5(3 + \sqrt{21})}{5\times 4} = \frac{3 + \sqrt{21}}{4}$

as given in post #2.
• Jan 5th 2012, 03:08 PM
ashleysmithd
Re: Help with Surds
Ahh.. I understand now. It's just simplifying the fraction ultimately.

Many thanks pickslides and earboth.