# [SOLVED] calc? ulate this fractions

• Aug 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
• Aug 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: $\displaystyle \frac{x^3-9x}{x^2-7x+12}$.

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

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

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

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

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

$\displaystyle x^2-7x+12$,

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

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

$\displaystyle \frac{x^3-9x}{x^2-7x+12} = \frac{x(x^2-9)}{x^2-7x+12}$
$\displaystyle =\frac{x(x+3)(x-3)}{x^2-7x+12} $$\displaystyle =\frac{x(x+3)(x-3)}{(x-3)(x-4)} . Finally cancel common terms to get: \displaystyle \frac{x^3-9x}{x^2-7x+12} = \frac{x(x^2-9)}{x^2-7x+12} \displaystyle =\frac{x(x+3)(x-3)}{x^2-7x+12}$$\displaystyle =\frac{x(x+3)(x-3)}{(x-3)(x-4)} $$\displaystyle =\frac{x(x+3)}{x-4} . RonL • Aug 28th 2006, 08:44 AM CaptainBlack Don't make duplicate posts in a number of forums, it confuses and wastes the time of the helpers who cannot tell if a question has been answered elsewhere. Other thread deleted. RonL • Aug 29th 2006, 05:05 AM CaptainBlack Quote: Originally Posted by asdfghjkl For some of my math i just cant figure out what i have to do just to start Any help or review to show me how to do some would be great? here is a link to the questions http://www.math.unb.ca/ready/paper/ 2a. Rationalise the denominator of: \displaystyle \frac{2}{\sqrt{3}+\sqrt{2}}\ \ \ \ ...(1) Now remember what we said earlier about the difference od two squares: \displaystyle a^2-b^2=(a+b)(a-b) If we put \displaystyle a=\sqrt{3} and \displaystyle b=\sqrt{2}, we have: \displaystyle 3-2=(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2}) So if we now multiply the top and bottom of \displaystyle (1) by \displaystyle \sqrt{3}-\sqrt{2} we don't change its value so: \displaystyle \frac{2}{\sqrt{3}+\sqrt{2}}=\frac{2(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$$\displaystyle =\frac{2\sqrt{3}-2\sqrt{2}}{3-2}=2\sqrt{3}-2\sqrt{2}$

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

$\displaystyle 5^{x+1}=25$

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

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

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

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

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

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

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

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

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

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

and:

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

Now:

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

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

RonL