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xEnOnn
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How do I solve for t in the equation \[\frac{t(t^22)}{t^21}=0\]
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
xEnOnn Group Title
How do I solve for t in the equation \[\frac{t(t^22)}{t^21}=0\] 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

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snoopshoag Group TitleBest ResponseYou've already chosen the best response.0
Since the equation is equal to zero, only the numerator is important.
 2 years ago

xEnOnn Group TitleBest ResponseYou've already chosen the best response.0
Why only the numerator is important?
 2 years ago

snoopshoag Group TitleBest ResponseYou've already chosen the best response.0
If t=0, the entire equation will equal 0, likewise, if t^22=0 the entire equation will equal 0.
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.2
If A/B = 0, then multiply both sides by B to get A = B*0 > A = 0
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.2
that's why only the numerator matters
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.2
besides, you can't divide by zero
 2 years ago

xEnOnn Group TitleBest ResponseYou've already chosen the best response.0
don't I have to check for \[t^21 \neq 0\]?
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.2
yes it's best to find out which numbers make the denominator zero first
 2 years ago

xEnOnn Group TitleBest ResponseYou've already chosen the best response.0
That would give t=1 or t=1. Then the solution to t would t=+sqrt(2), t=0 and t not equals +1?
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.2
With the solution, you only need to worry about what t equals (instead of what it can't equal)
 2 years ago

snoopshoag Group TitleBest ResponseYou've already chosen the best response.0
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.
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.2
type \frac{x}{y} to type out \[\Large \frac{x}{y}\]
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.2
for some reason, this is missing in the equation editor...
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.2
np
 2 years ago

xEnOnn Group TitleBest ResponseYou've already chosen the best response.0
Suppose another function call it E(t) holds only if the above equation of \[\frac{t(t^22)}{t^21} \neq 0\], 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?
 2 years ago

xEnOnn Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

jim_thompson5910 Group TitleBest ResponseYou've already chosen the best response.2
yes that possibility exists, but you just ignore those potential solutions if it arises
 2 years ago

snoopshoag Group TitleBest ResponseYou've already chosen the best response.0
Correct, the function would not work if \[t=\pm1\], one way to avoid this would be to define the domain, so t=\[(\infty,1),(1,1),(1,\infty)\]
 2 years ago

xEnOnn Group TitleBest ResponseYou've already chosen the best response.0
Thanks everyone! :)
 2 years ago

snoopshoag Group TitleBest ResponseYou've already chosen the best response.0
You're welcome.
 2 years ago
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