## xEnOnn 4 years ago 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!

1. snoopshoag

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

2. xEnOnn

Why only the numerator is important?

3. snoopshoag

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

4. jim_thompson5910

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

5. jim_thompson5910

that's why only the numerator matters

6. jim_thompson5910

besides, you can't divide by zero

7. xEnOnn

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

8. jim_thompson5910

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

9. xEnOnn

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

10. jim_thompson5910

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

11. snoopshoag

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.

12. jim_thompson5910

type \frac{x}{y} to type out $\Large \frac{x}{y}$

13. jim_thompson5910

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

14. snoopshoag

Thanks!

15. jim_thompson5910

np

16. xEnOnn

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?

17. xEnOnn

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?

18. jim_thompson5910

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

19. snoopshoag

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

20. xEnOnn

Thanks everyone! :)

21. snoopshoag

You're welcome.