Rearranging an equationDate: 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 - 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. If you'd like more information on how to solve cubic equations 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}} (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 |
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