Solve 2

[ashmarissa]

t^2 + 1 = 13/6t

 

[stapel]

t^2 + 1 = 13/6t

What you have posted could be interpreted as meaning the following:

. . . . .\(\displaystyle t^2 + 1 = \frac{13}{6t}\)

Is this what you meant?

When you reply, please include a clear listing of your steps and reasoning so far. Thank you! :wink:

 

[ashmarissa]

Yes that is what I meant. So far I have moved the 13/6 to the left side. I think the fraction has me stumped.

 

[stapel]

Yes that is what I meant. So far I have moved the 13/6 to the left side. I think the fraction has me stumped.

What "13/6"? Either you have a "13/(6t)" or a "(13/6)t", but, according to what you've posted, you have no "13/6" term.

Please reply showing your steps. Thank you! :wink:

 

[ashmarissa]

Started with t^2 + 1= 13/6t
I moved the 13/6t to the left side.
I now have t^2 + 13/6t +1=0.
After that, I feel lost.

 

[Deleted member 4993]

Started with t^2 + 1= 13/6t
I moved the 13/6t to the left side.
I now have t^2 + 13/6t +1=0.
After that, I feel lost.

I want to make sure:

The original term on the right-hand-side:

Was it \(\displaystyle \dfrac{13}{6t}\), or,

Was it \(\displaystyle \dfrac{13}{6}t\)

This is very important factor in choosing the method of solution.

 

[HallsofIvy]

If this were \(\displaystyle t^2+ 1= (13/6)t\) then it would be equivalent to \(\displaystyle t^2- (13/6)t+ 1= 0\) which could be solved using the quadratic formula.

If you mean \(\displaystyle t^2+ 1= 13/(6t)\) then the first thing I would do is multiply both sides by 6t to eliminate the fraction:
\(\displaystyle 6t^3+ 6t= 13\) which is equivalent to the cubic equation \(\displaystyle 6t^3+ 6t- 13= 0\). You can use the "rational root theorem" to show that the only possible rational roots are 13, -13, 13/2, -13/2, 13/3, -13/3, 13/6, and -13/6.

However, evaluating \(\displaystyle 6t^3+ 6t- 13\) at each of those shows that they do NOT satisfy the equation. There are no rational numbers satisfying the equation so you will have to use the "cubic formula":http://en.wikipedia.org/wiki/Cubic_function#Cardano.27s_method

That's much more complicated than the quadratic formula!