The Math Forum

Ask Dr. Math - Questions and Answers from our Archives
Associated Topics || Dr. Math Home || Search Dr. Math

Rearranging an equation

Date: 15 Mar 1995 12:48:40 -0500
From: Lee Carkenord
Subject: equation help

I have an equation that need to be re-arranged. It seems so 
simple but I can't do it. Could you please help?  If:
           ___              ___
          |  V*V      V*V*V    |
     X=Y* |------- + ----------|
          | 3481      148485   |
          |___               __|
I need this equation re-arranged to solve for V.    That is, 
V=????  Oh, the symbol    *    means multiply or "times".
                        Thank you,  Lee in Denver

Date: 15 Mar 1995 14:39:58 -0500
From: Dr. Ken
Subject: Re: equation help

Hello there.

In fact, although it may seem simple, this is actually quite 
hard.  See, what you've got here is a cubic equation (as 
opposed to a quadratic or linear equation), since the power 
of v is 3.  These are notoriously difficult to solve.  I fed it 
into a computer which runs Mathematica, and it gave me the 
following answers (there are 3):

{{v -> -49495/3481 + (49495*4899510050^(1/3)*y)/
        (3481*(126541600923*x*y^2 - 4899510050*y^3 +  
         205379*3^(1/2)*(x*(126541600923*x - 
         9799020100*y)*y^4)^(1/2))^(1/3)) + ((49495/2)^(1/3)*
         (126541600923*x*y^2 - 4899510050*y^3 + 
         205379*3^(1/2)*(x*(126541600923*x - 	
{v -> -49495/3481 - (49495*(49495/2)^(2/3)*
        (1 + I*3^(1/2))*y)/(3481*(126541600923*x*y^2 - 
        4899510050*y^3 + 205379*3^(1/2)*(x*
        (126541600923*x - 9799020100*y)*y^4)^(1/2))^
        (1/3)) - ((49495/2)^(1/3)*(1 - I*3^(1/2))*
        (126541600923*x*y^2 - 4899510050*y^3 + 
        205379*3^(1/2)*(x*(126541600923*x - 

{v -> -49495/3481 + ((49495*I)/3481*(49495/2)^(2/3)*(I + 
        3^(1/2))*y)/(126541600923*x*y^2 - 4899510050*y^3 + 
        205379*3^(1/2)*(x*(126541600923*x - 9799020100*
        y)*y^4)^(1/2))^(1/3) /- ((49495/2)^(1/3)*(1 + I*3^(1/2))*
        (126541600923*x*y^2 - 4899510050*y^3 + 
         205379*3^(1/2)*(x*(126541600923*x - 9799020100*

Now, one might assume that you're only looking for the real 
answer (as opposed to complex), in which case look only at t
he first one.  One complication: the other roots might end up 
being real, depending on what x and y are.  To see this, let 
x = 0.05, and let y = 1.  Then graph v^2/3481 + v^3/148485, 
and notice that there are three places where this graph reaches the 
height 0.05.  But if x=1 and y=1, there is only one real solution, 
and the other 2 are complex.

I hope this answer is useful to you, but I suspect that it might 
just be unsightly enough to convince you that it's best to leave 
the equation as it's written.  

If you'd like more information on how to solve cubic equations 
by hand, ask us, or check out our web page:   

-Ken "Dr." Math

Date: 15 Mar 1995 22:47:10 -0500
From: Lee Carkenord
Subject: Re: equation help

Hey, Ken---that's great!! Thanks a bunch. That "Mathmatica" 
must be something.  Now--will that long equation actually 
solve a problem???  
        If X = 9,  Y = 3,  what is V?
        If X = 29.626,  and Y = 3.442,  what is V?  
Can you do that for me??
                        Lee in Denver

Date: 17 Mar 1995 20:17:57 -0500
From: Dr. Ken
Subject: Re: equation help

Hello there!

I plugged in your numbers to Mathematica, and here's 
what I got:

In[3]:= N[Solve[9==3*(v^2 / 3481 + v^3 / 148485), v]]

Out[3]= {{v -> 64.4811}, {v -> -53.5685 + 63.551 I},
               {v -> -53.5685 - 63.551 I}}

(these are the three possible answers for your first set of 
numbers, and one answer is real)

In[4]:= N[Solve[29.626==3.442*(v^2 / 3481 +
            v^3 / 148485), v]]

Out[4]= {{v -> -69.3306 - 92.2251 I}, {v -> -69.3306 + 
              92.2251 I},
                {v -> 96.0053}}

(these are the answers for the second set of numbers, and again 
only one answer is real).

I actually plugged your numbers into the original equation, and 
then asked Mathematica to solve the resulting equation.  It was 
much less typing than plugging into the formulas that I gave you.

Incidentally, if you want to find out for which values of X and 
Y you'd have to worry about there being multiple roots of the 
equation, you could use Calculus to find the Min and Max of 
the right-hand side (by taking the derivative with respect to V, 
and setting it equal to 0) and then figuring out whether the X 
value falls within the triple-value range: the range between the 
Min and the Max.

Then if you're armed with this information and you're trying 
to solve the equation for a particular value of X and Y, you 
might only have to look at the real-valued solution to the 
equation, and not worry about those other two.

I hope this helps.

-Ken "Dr." Math
Associated Topics:
High School Basic Algebra

Search the Dr. Math Library:

Find items containing (put spaces between keywords):
Click only once for faster results:

[ Choose "whole words" when searching for a word like age.]

all keywords, in any order at least one, that exact phrase
parts of words whole words

Submit your own question to Dr. Math

[Privacy Policy] [Terms of Use]

Math Forum Home || Math Library || Quick Reference || Math Forum Search

Ask Dr. MathTM
© 1994- The Math Forum at NCTM. All rights reserved.