## 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! 3 years ago 3 years ago

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.