## xEnOnn Group Title 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! 2 years ago 2 years ago

1. snoopshoag Group Title

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

2. xEnOnn Group Title

Why only the numerator is important?

3. snoopshoag Group Title

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

4. jim_thompson5910 Group Title

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

5. jim_thompson5910 Group Title

that's why only the numerator matters

6. jim_thompson5910 Group Title

besides, you can't divide by zero

7. xEnOnn Group Title

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

8. jim_thompson5910 Group Title

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

9. xEnOnn Group Title

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 Group Title

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

11. snoopshoag Group Title

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 Group Title

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

13. jim_thompson5910 Group Title

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

14. snoopshoag Group Title

Thanks!

15. jim_thompson5910 Group Title

np

16. xEnOnn Group Title

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 Group Title

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 Group Title

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

19. snoopshoag Group Title

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 Group Title

Thanks everyone! :)

21. snoopshoag Group Title

You're welcome.