# [SOLVED] calc? ulate this fractions

• August 28th 2006, 08:15 AM
asdfghjkl
[SOLVED] calc? ulate this fractions
For some of my math i just cant figure out what i have to do just to start
Any help or review to show me how to do some would be great?
here is a link to the questions
• August 28th 2006, 08:30 AM
CaptainBlack
Quote:

Originally Posted by asdfghjkl
For some of my math i just cant figure out what i have to do just to start
Any help or review to show me how to do some would be great?
here is a link to the questions

There is a lot here, and it would be a mistake for us to do them all for you.
But I see no reason for not showing you how to do a selection of these.

1a. Simplify: $\frac{x^3-9x}{x^2-7x+12}$.

First if $x$ is a common factor of the top or bottom of a
rational expression like this, factor it out:

$\frac{x^3-9x}{x^2-7x+12} = \frac{x(x^2-9)}{x^2-7x+12}$.

Now look for terms which are the difference of two squares as we know
that $a^2+b^2=(a+b)(a-b)$, here we have such a term, so:

$
\frac{x^3-9x}{x^2-7x+12} = \frac{x(x^2-9)}{x^2-7x+12}$

$
=\frac{x(x+3)(x-3)}{x^2-7x+12}
$

Now we check if we can cancel any terms, in this case we check if:

$x^2-7x+12$,

has one or more of $x$, $x+3$ or $x-3$
as factors. The first obviously is not a factor, if the second was a factor
then substituting $x=-3$ into the expression would simplify to 0,
it does not so $x+3$ id not a factor of $x^2-7x+12$.
Now if the last candidate is a factor substituting $x=3$ into the
expressions would simplify to zero, which in this case it does, and by trial
and error we find:

$x^2-7x+12=(x-3)(x-4)$, so now we have:

$
\frac{x^3-9x}{x^2-7x+12} = \frac{x(x^2-9)}{x^2-7x+12}$

$
=\frac{x(x+3)(x-3)}{x^2-7x+12}
$
$
=\frac{x(x+3)(x-3)}{(x-3)(x-4)}
$
.

Finally cancel common terms to get:

$
\frac{x^3-9x}{x^2-7x+12} = \frac{x(x^2-9)}{x^2-7x+12}$

$
=\frac{x(x+3)(x-3)}{x^2-7x+12}
$
$
=\frac{x(x+3)(x-3)}{(x-3)(x-4)}
$
$
=\frac{x(x+3)}{x-4}
$
.

RonL
• August 28th 2006, 08:44 AM
CaptainBlack
Don't make duplicate posts in a number of forums, it confuses and wastes the
time of the helpers who cannot tell if a question has been answered elsewhere.

RonL
• August 29th 2006, 05:05 AM
CaptainBlack
Quote:

Originally Posted by asdfghjkl
For some of my math i just cant figure out what i have to do just to start
Any help or review to show me how to do some would be great?
here is a link to the questions

2a. Rationalise the denominator of:

$
\frac{2}{\sqrt{3}+\sqrt{2}}\ \ \ \ ...(1)
$

Now remember what we said earlier about the difference od two squares:

$
a^2-b^2=(a+b)(a-b)
$

If we put $a=\sqrt{3}$ and $b=\sqrt{2}$, we have:

$
3-2=(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})
$

So if we now multiply the top and bottom of $(1)$ by $\sqrt{3}-\sqrt{2}$ we don't change its value so:

$
\frac{2}{\sqrt{3}+\sqrt{2}}=\frac{2(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$
$
=\frac{2\sqrt{3}-2\sqrt{2}}{3-2}=2\sqrt{3}-2\sqrt{2}
$

RonL
• September 2nd 2006, 03:59 AM
CaptainBlack
4a. Solve for x (without using a claculator)

$
5^{x+1}=25
$

$25=5\times 5=5^2$, so:

$
5^{x+1}=5^2
$
.

Therefore equating exponents $x+1=2$, and so $x=1$.

Can also be done by taking logs to the base 5:

$
(x+1)\log_5(5)=2\log_5(5)
$

but $\log_5(5)=1$, so we are back to where we were, and $x=1$

RonL
• September 2nd 2006, 04:17 AM
CaptainBlack
5a. Simplify:

$
\log_2(5)+\log_2(x^2-1)-\log_2(x-1)
$
.

Here the first things we need are some of the basic laws of logarithms:

$\log_a(u)+\log_s(v)=\log_a(u\times v)$,

and:

$\log_a(u)-\log_s(v)=\log_a(u/v)$.

Now:

$
\log_2(5)+\log_2(x^2-1)$
$-\log_2(x-1)=(\log_2(5)+\log_2(x^2-1))-\log_2(x-1)
$

$
=\log_2(5(x^2-1))-\log_2(x-1)
$
$
=\log \left(\frac{5(x^2-1}{x-1}\right)
$

Finally we need to recall how to factorise the difference of two squares to get:

$
=\log \left(\frac{5(x-1)(x+1)}{x-1}\right)=\log(5(x+1))
$

RonL