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/

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- Aug 28th 2006, 08:15 AMasdfghjkl[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

http://www.math.unb.ca/ready/paper/ - Aug 28th 2006, 08:30 AMCaptainBlackQuote:

Originally Posted by**asdfghjkl**

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 AMCaptainBlack
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 AMCaptainBlackQuote:

Originally Posted by**asdfghjkl**

$\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 AMCaptainBlack
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 AMCaptainBlack
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