Solve 3b^(2/3) - 2b = 10/16

[dlp211]

The problem statement, all variables and given/known data

3b^(2/3) - 2b = 10/16

Find solution to a precision of thousanths



The attempt at a solution

I know the answer is ~ .199, I do not however know how to actually solve the above equation. If anyone can help, I'd appreciate it, it's been over 10 years since I took an algebra class.

[SammyS]

The problem statement, all variables and given/known data

3b^(2/3) - 2b = 10/16

Find solution to a precision of thousandths

The attempt at a solution

I know the answer is ~ .199, I do not however know how to actually solve the above equation. If anyone can help, I'd appreciate it, it's been over 10 years since I took an algebra class.
Reduce 10/16 → 5/8.

Isolate the 3b2/3 by adding 2b to both sides.

Cube both sides.

The result is a cubic equation. This particular cubic equation does not have any rational solutions.

Use the bisection method, starting with a very small interval near 0.2. There is the solution near 0.2, as you said, but there's another nearby; between -0.1 and 0 .

There is a third solution that's between 2 and 3.

[kunguz]

The problem statement, all variables and given/known data

3b^(2/3) - 2b = 10/16

Find solution to a precision of thousanths



The attempt at a solution

I know the answer is ~ .199, I do not however know how to actually solve the above equation. If anyone can help, I'd appreciate it, it's been over 10 years since I took an algebra class.

I believe Newton-Rhapsody iterative solving method might help you to find the roots if they are real ofcourse:

http://en.wikipedia.org/wiki/Newton%27s_method

[dlp211]

Thanks guys,

This was the end of a much larger problem and it now makes sense that I need to use the Newton's Method(this is was the end of a Calc II) problem.

It's been a long time for some algebra concepts for me so I appreciate all the help.

Dave

[SammyS]

Thanks guys,

This was the end of a much larger problem and it now makes sense that I need to use the Newton's Method(this is was the end of a Calc II) problem.

It's been a long time for some algebra concepts for me so I appreciate all the help.

Dave
This is the pre-calculus section, so I suggested bisection.

Make sure your initial guess is close enough to the root you want to find. Newton's Method may find a different root if you don't start close enough to the one you're interested in.

[Mark44]

I believe Newton-Rhapsody iterative solving method might help you to find the roots if they are real ofcourse:

http://en.wikipedia.org/wiki/Newton%27s_method

There's Newton-Raphson, which is probably what you were thinking of.

[HallsofIvy]

But "Rhapsody" sounds so much better!

[Mark44]

There's Newton-Raphson, which is probably what you were thinking of.

But "Rhapsody" sounds so much better!

Well, there's that.