Solving a Cubic: Help Appreciated!

[pita0001]

I'm trying to solve the cubic:

2t^3=5t-11t^2

Been stuck on this for awhile. Any help is appreciated.

First I took everything over to one side, so.. 2t^3-5t+11t^2
then set it to zero 2t^3-5t+11t^2=0
then didivded by t..so t(2t^2-5+11t)

Then I tried multiplying 11 by 2 which =22
but 22 and 5 don't factor.Any help is greatly appreciated. Thanks!

 

[MarkFL]

Hello and welcome to MHB! :D

I edited your post to include the problem statement...it is okay to put the problem in the title, but we ask that you also include it in your post as well for clarity.

We are given to solve:

\(\displaystyle 2t^3=5t-11t^2\)

I would arrange it in standard form (descending exponents) as:

\(\displaystyle 2t^3+11t^2-5t=0\)

Factor out the $t$:

\(\displaystyle t\left(2t^2+11t-5 \right)=0\)

So, using the zero factor property, we equate each factor in turn to zero:

\(\displaystyle t=0\)

\(\displaystyle 2t^2+11t-5=0\)

Now, the linear factor gives us the root $t=0$, but the quadratic factor does not further factor in the conventional sense (check the discriminant and see that it is not a perfect square). How else can we solve quadratics besides factoring?

 

[pita0001]

What about doing the quadratic formula would that work? (To further solve it) or no?

 

[MarkFL]

What about doing the quadratic formula would that work? (To further solve it) or no?

Yes, that would work...the quadratic formula is a general formula that works for all quadratics. Another option is completing the square.

 

[pita0001]

So

t=-11+- (square root) 161/4that would be my final answer, right?

 

[MarkFL]

So

t=-11+- (square root) 161/4that would be my final answer, right?

If you mean:

t = (-11 ± sqrt(161))/4

Then yes, these are the roots to the quadratic factor. But don't forget you also have t = 0 as the root from the other factor. So, you have the following 3 solutions:

\(\displaystyle t=0,\,\frac{-11\pm\sqrt{161}}{4}\)

 

[pita0001]

Where does t=0 come from?

 

[MarkFL]

Where does t=0 come from?

Recall we originally factored the equation as:

\(\displaystyle t\left(2t^2+11t-5 \right)=0\)

So, we need to equate each factor to zero to find all of the roots:

\(\displaystyle t=0\)

\(\displaystyle 2t^2+11t-5=0\)

The first equation gives us \(\displaystyle t=0\) and the second gives us \(\displaystyle t=\frac{-11\pm\sqrt{161}}{4}\), for a total of 3 solutions.

 

[pita0001]

Ah, right!
Gratzi, gratzi!