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xEnOnn

  • 3 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!

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  1. snoopshoag
    • 3 years ago
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    Since the equation is equal to zero, only the numerator is important.

  2. xEnOnn
    • 3 years ago
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    Why only the numerator is important?

  3. snoopshoag
    • 3 years ago
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    If t=0, the entire equation will equal 0, likewise, if t^2-2=0 the entire equation will equal 0.

  4. jim_thompson5910
    • 3 years ago
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    If A/B = 0, then multiply both sides by B to get A = B*0 ---> A = 0

  5. jim_thompson5910
    • 3 years ago
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    that's why only the numerator matters

  6. jim_thompson5910
    • 3 years ago
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    besides, you can't divide by zero

  7. xEnOnn
    • 3 years ago
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    don't I have to check for \[t^2-1 \neq 0\]?

  8. jim_thompson5910
    • 3 years ago
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    yes it's best to find out which numbers make the denominator zero first

  9. xEnOnn
    • 3 years ago
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    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
    • 3 years ago
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    With the solution, you only need to worry about what t equals (instead of what it can't equal)

  11. snoopshoag
    • 3 years ago
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    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
    • 3 years ago
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    type \frac{x}{y} to type out \[\Large \frac{x}{y}\]

  13. jim_thompson5910
    • 3 years ago
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    for some reason, this is missing in the equation editor...

  14. snoopshoag
    • 3 years ago
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    Thanks!

  15. jim_thompson5910
    • 3 years ago
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    np

  16. xEnOnn
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    yes that possibility exists, but you just ignore those potential solutions if it arises

  19. snoopshoag
    • 3 years ago
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    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
    • 3 years ago
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    Thanks everyone! :)

  21. snoopshoag
    • 3 years ago
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    You're welcome.

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