# Tricky equation to solve

tedwillis
Hi,

I need to solve this equation that stems from a differential equation, though that isn't really important. Here's the equation that needs to be solved for t:

3t+100=4((9t^3+900t^2+30000t)/(3t+100)…

I've tried myself and always seem to hit a dead end. Using a calculator, I know the answer is 100/3(2^2/3-1) or ~19.58, but can't find a way to get to this answer.

I should mention, I would like this in a fully worked solution if possible. The subject does not allow calculators.

Homework Helper
3t+100=4((9t^3+900t^2+30000t)/(3t+100)…
The three dots at the end of an equation usually means "and so on" - meaning to continue some sort of pattern. Is that what you meant here?
(there's also an extra bracket in there.)

Guessing "no" ... rewriting - you need t to satisfy the relation: $$(3t+100)^2 = 4(9t^3+900t^2+30000t)$$your first task (brute force) is to put the equation into standard form.

From the look of it, it is a cubic, so as many as three values of t may make this equation true. You should already know how to solve for the roots of a cubic.

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tedwillis
Sorry, wrote the question wrong:
3t+100=4((9t^3+900t^2+30000t)/(3t+100)^2)

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tedwillis

x(t) is the amount of an element in a container

The original d.e:
$$dx/dt+6x/(100+3t)=3$$

100+3t=v, the total volume of liquid in a container over time. The max of this is 300.

Solving the d.e gives:
$$x(t)=(9t^3+900t^2+30000t)/(3t+100)^2)$$

Now, and this is where I am stuck, I have to find when the percentage of the element within the total liquid is ≥ 25%. This occurs when $$x/v=1/4$$, so $$3t+100=4x$$, hence:
$$3t+100=4((9t^3+900t^2+30000t)/(3t+100)^2)$$

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Homework Helper
Hmmm I get ##x(t)= c(3t-100)^2-3t+100## where c is the constant of integration.

But if you are still happy with yours - you still have a cubic.
Put it in standard form.

tedwillis
Hmmm I get ##x(t)= c(3t-100)^2-3t+100## where c is the constant of integration.

But if you are still happy with yours - you still have a cubic.
Put it in standard form.

Oh sorry, I forgot to mention the intial vlaue is x(0)=0

tedwillis
So I guess this is what you mean by standard form:

$$9t^3+900t^2+30000t-1000000=0$$

Homework Helper
Dearly Missed
Sorry, wrote the question wrong:
3t+100=4((9t^3+900t^2+30000t)/(3t+100)^2)

It is easier (and more insightful) to re-scale this: set t = 100s in the equation, to get
$$(3s+1)^3 = 12(3s^3 + 3s^2 + s).$$ It is easier to get high accuracy on a calculator when you do this, plus it is easier to manipulate, etc.

RGV

Staff Emeritus
Homework Helper
Gold Member
It is easier (and more insightful) to re-scale this: set t = 100s in the equation, to get
$$(3s+1)^3 = 12(3s^3 + 3s^2 + s).$$ It is easier to get high accuracy on a calculator when you do this, plus it is easier to manipulate, etc.

RGV
Following Ray's suggestion, let u = 3s+1. so that $\displaystyle s=\frac{u-1}{3}\ .$

That should make the algebra even cleaner.