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
___              ___
|  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

{{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 -
9799020100*y)*y^4)^(1/2))^(1/3))/(3481*y)},

{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 -
9799020100*y)*y^4)^(1/2))^(1/3))/(6962*y)},

{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*
y)*y^4)^(1/2))^(1/3))/(6962*y)}}

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.

by hand, ask us, or check out our web page:
http://mathforum.org/dr.math/dr-math.html

-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}}

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

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