# Annoying Algebra/Indices question

• Jan 10th 2009, 04:49 PM
MellowCandle
Annoying Algebra/Indices question
Hello there!

I found a Indices/Algebra question which I think the writer has answered incorrectly. It goes like this;

http://homepage.ntlworld.com/james.m...math/alg_q.gif

On the step from last, they get a value of -2 and a positive + t ^ -2. I don't see where this has come from. When I went through this myself, I got a final answer of;

t^2 - (1 / (t^2)) + 2

who is correct?
• Jan 10th 2009, 05:14 PM
Jhevon
Quote:

Originally Posted by MellowCandle
Hello there!

I found a Indices/Algebra question which I think the writer has answered incorrectly. It goes like this;

http://homepage.ntlworld.com/james.m...math/alg_q.gif

On the step from last, they get a value of -2 and a positive + t ^ -2. I don't see where this has come from. When I went through this myself, I got a final answer of;

t^2 - (1 / (t^2)) + 2

who is correct?

they are right.

what you did was this: $(a + b)^2 = a^2 + b^2$

this is very wrong, and you should never do that again

remember, $(a \pm b)^2 = a^2 \pm 2ab + b^2$

thus: $(t - t^{-1})^2 = t^2 - 2t(t^{-1}) + (t^{-1})^2 = t^2 - 2 + t^{-2}$
• Jan 10th 2009, 05:53 PM
MellowCandle
:o(
I'm having difficulty understanding the last calculations.

I've broken;

( t - t^-1 ) ^ 2

into;

( t - t^-1 ) ( t - t^-1 )

I've got t ^ 2, but can you explain how you go about solving;

t x -t ^ -1

and;

- t^-1 x - t^-1
• Jan 10th 2009, 06:15 PM
Jhevon
Quote:

Originally Posted by MellowCandle
I'm having difficulty understanding the last calculations.

I've broken;

( t - t^-1 ) ^ 2

into;

( t - t^-1 ) ( t - t^-1 )

I've got t ^ 2, but can you explain how you go about solving;

t x -t ^ -1

and;

- t^-1 x - t^-1

just follow the formula i gave you.

if you want to do it the long way, this is how

$(t - t^{-1})^2 = (t - t^{-1})(t - t^{-1})$

$= t^2 - (t \cdot t^{-1}) - (t^{-1} \cdot t) + (t^{-1})^2$ .........hopefully you see how i got to this line. i just expanded the brackets.

Now recall, when multiplying numbers of the same base, we add the indices. that is, $x^a \cdot x^b = x^{a + b}$

that means $t \cdot t^{-1} = t^{1 + (-1)} = t^0 = 1$

so we have

$t^2 - 1 - 1 + (t^{-1})^2$

$= t^2 - 2 + t^{-2}$

( $(t^{-1})^2$ becomes $t^{-2}$ since $(x^a)^b = x^{ab}$)
• Jan 10th 2009, 06:52 PM
MellowCandle
Hello,

Thanks for that, I think I've got it. I think I became phased by the presence of the minus signs in the following calculations;

$t \cdot - t^{-1}$

and

$- t^{-1} \cdot - t^{-1}$

It feels as if the negativity / positivity of the terms is treated 'crudely' (apologies for the poor description.) For example, in the first, the minus sign of the second term is nudged out of the way in order to make the calculation;

$- t^{1} \cdot t^{-1}$
• Jan 10th 2009, 06:59 PM
Jhevon
Quote:

Originally Posted by MellowCandle
Hello,

Thanks for that, I think I've got it. I think I became phased by the presence of the minus signs in the following calculations;

$t \cdot - t^{-1}$

and

$- t^{-1} \cdot - t^{-1}$

It feels as if the negativity / positivity of the terms is treated 'crudely' (apologies for the poor description.) For example, in the first, the minus sign of the second term is nudged out of the way in order to make the calculation;

$- t^{1} \cdot t^{-1}$

i'm sorry, i don't understand what you are saying.

here's what you need to know:

- a positive number times a positive number is positive
- a negative number times a negative number is positive
- a positive number times a negative number is negative
• Jan 10th 2009, 07:19 PM
MellowCandle
Thanks very much for all your help. I did the following from scratch on a notepad;

$t^{-1} \cdot - t^{-1}$

$= - (t^{-1}) ^{2}$

$= - t^{-1 \cdot 2}$

$= - t^{-2}$

I hope this is correct! (Thinking)
• Jan 10th 2009, 07:32 PM
Jhevon
Quote:

Originally Posted by MellowCandle
Thanks very much for all your help. I did the following from scratch on a notepad;

$t^{-1} \cdot - t^{-1}$

$= - (t^{-1}) ^{2}$

$= - t^{-1 \cdot 2}$

$= - t^{-2}$

I hope this is correct! (Thinking)

well, that is correct. but that's not what you have in this problem

you have

$-t^{-1} \cdot - t^{-1}$

$= {\color{red}+}~ (t^{-1}) ^{2}$ .....since negative times negative makes positive. or you can think of the + sign coming from squaring in this particular case, since $(-t^{-1})^2 = + t^{-2}$

$= t^{-1 \cdot 2}$

$= t^{-2}$

that's why they have a positive $t^{-2}$