Question

How do I solve for t in the equation \[\frac{t(t^2-2)}{t^2-1}=0\]

I know the answers are +-square root of 2 and 0 but that's after I punch it into my calculator. Could you show me the steps if I want to evaluate this by hand?

Thanks!

Answer 1

Since the equation is equal to zero, only the numerator is important.

Answer 2

Why only the numerator is important?

Answer 3

If t=0, the entire equation will equal 0, likewise, if t^2-2=0 the entire equation will equal 0.

Answer 4

If A/B = 0, then multiply both sides by B to get A = B*0 ---> A = 0

Answer 5

that's why only the numerator matters

Answer 6

besides, you can't divide by zero

Answer 7

don't I have to check for \[t^2-1 \neq 0\]?

Answer 8

yes it's best to find out which numbers make the denominator zero first

Answer 9

That would give t=1 or t=-1. Then the solution to t would t=+-sqrt(2), t=0 and t not equals +-1?

Answer 10

With the solution, you only need to worry about what t equals (instead of what it can't equal)

Answer 11

Hey jim, sorry to butt in on this conversation, but do you know how to create a fraction in the equation editor? It is very difficult to give coherent answers without fractions.

Answer 12

type \frac{x}{y} to type out \[\Large \frac{x}{y}\]

Answer 13

for some reason, this is missing in the equation editor...

Answer 14

Thanks!

Answer 15

np

Answer 16

Suppose another function call it E(t) holds only if the above equation of \[\frac{t(t^2-2)}{t^2-1} \neq 0\],

Then, for E(t) to hold, t must be \[t \neq \pm \sqrt{2}\]
\[t \neq 0\]

But then what if the function was given -1 or 1 into it? The whole thing would not work since it will be division by zero, wouldn't it?

Answer 17

As in, since the condition that I have got from the equation, was just \[t \neq \pm \sqrt{2} \; \; \;
and \; \; \; t \neq 0\]

Then wouldn't there be a possibility/danger that someone throws in a value of t which is -1 or 1?

Answer 18

yes that possibility exists, but you just ignore those potential solutions if it arises

Answer 19

Correct, the function would not work if \[t=\pm1\], one way to avoid this would be to define the domain, so t=\[(-\infty,-1),(-1,1),(1,\infty)\]

Answer 20

Thanks everyone! :)

Answer 21

You're welcome.