Is there an explicit solution to the equation \sqrt{Ax-x^3}+\sqrt{Bx-x^3}=C?

[Apteronotus]

Solving a cubic... sort of!

Hi,

Can the equation \sqrt{Ax-x^3}+\sqrt{Bx-x^3}=C be solved explicitly?

All of MathLab, Maple and WolframAlpha seem to give an explicit solution but they don't show how they come to it. I'm afraid they may be missing other possible solutions.

Thanks,

 

[lanedance]

why not start by simplifying to
(\sqrt{A}+\sqrt{B})\sqrt{x-x^3} = C

 

[lanedance]

note that if you square things you may need to account for the fact the argument of a square root cannot be zero

 

[aegrisomnia]

why not start by simplifying to

Well, for starters, your factorization is incorrect...

 

[lanedance]

good point, read it incorrectly, must be late ;)... will have another look

 

[lanedance]

so i did a wolfram alpha check and the solution looks nasty... so there probably is a closed form method but its likely pretty involved & tedious, after looking at it closer, I can't see any easy way to simplify...

 

[lanedance]

getting late, so I've probably made a mistake, but see what you think of this logic...
assume x>0 and hopefully we don't divide by zero anywhere
\sqrt{x}(\sqrt{A-x^2}+\sqrt{B-x^2})=C
multiply by the rational to get
\sqrt{x}(A-B)=C(\sqrt{A-x^2}-\sqrt{B-x^2})

re-arrange both to get
\sqrt{A-x^2}+\sqrt{B-x^2}=\frac{C}{\sqrt{x}}
\sqrt{A-x^2}-\sqrt{B-x^2}=\frac{\sqrt{x}(A-B)}{C}

adding and subtracting we get to
2\sqrt{A-x^2}=\frac{C}{\sqrt{x}}+\frac{\sqrt{x}(A-B)}{C}
2\sqrt{B-x^2}= \frac{C}{\sqrt{x}}-\frac{\sqrt{x}(A-B)}{C}

square both
4(A-x^2)=\frac{C^2}{x}+2(A-B)+\frac{x(A-B)^2}{C^2}
2(B-x^2)= \frac{C^2}{x}-2(A-B)+\frac{x(A-B)^2}{C^2}

which is starting to look a bit more tractable as in effect we can solve a cubic now...

 

[aegrisomnia]

Here's another thought...

In the original equation, solutions exist only for positive C, and both sides of the equation are trivially positive where solutions exist (in the real numbers).

Given this, square both sides of the equation; since if f(x) = g(x) and both f(x) and g(x) are positive, then [f(x)]^2 = [g(x)]^2 iff f(x) = g(x).

Rearrange terms so that only the radical term (only one remains after squaring) is on the right, and everything else is on the left of the equation. Clearly the right-hand side is positive, so solutions exist only when the left-hand side is positive... we must check our answers for x to ensure that this condition is satisfied. For now, square again.

Combine all like terms, and you arrive (unless I goofed) at an equation like this:

(4C^2)x^3 + (A^2 + B^2 - 2AB)x^2 - 2C^2(A + B)x + C^4 = 0.

This is a cubic equation in standard form, and since there is a method to solve cubics, this can be solved. Simply check the answers against the conditions we have identified and that should contain all answers...

 

[disregardthat]

Hi,

Can the equation \sqrt{Ax-x^3}+\sqrt{Bx-x^3}=C be solved explicitly?

\sqrt{Ax-x^3}+\sqrt{Bx-x^3}=C

so

C(\sqrt{Ax-x^3}-\sqrt{Bx-x^3}) = Ax-x^3-(Bx-x^3) = (A-B)x

from the original equation

\sqrt{Bx-x^3}=C-\sqrt{Ax-x^3}

which we can insert into our new equation:

(A-B)x = C(\sqrt{Ax-x^3}-(C-\sqrt{Ax-x^3})) = 2C\sqrt{Ax-x^3}-C^2

2C\sqrt{Ax-x^3} = (A-B)x+C^2

squaring yields

2C(Ax-x^3) = (A-B)^2x^2+2C^2(A-B)x+C^4

i.e.

2Cx^3+(A-B)^2x^2 + (2C^2(A-B) - 2AC)x+C^4 = 0

which can be solved using the formula for third-degree polynomials.

If you want to solve it yourself without the formula, I suggest the following:
- do a linear transformation of x to remove the coefficient of x^2
- substitute x = y+a/y for a suitable constant a to transform it into a second-degree polynomial equation in y^2

Note that we may have generated more solutions that there are by squaring, so make sure that you are finding the correct ones.

 

[Mute]

2C\sqrt{Ax-x^3} = (A-B)x+C^2

squaring yields

2C(Ax-x^3) = (A-B)^2x^2+2C^2(A-B)x+C^4

You forgot to square the 2C in going to the second line there.

 

[aegrisomnia]

^ Does that mean his reduction to a cubic yields the same thing that I said in the post before his?

 

[lanedance]

I think we've all done the same thing in essence - your post in #7 probably gives the clearest method

 

[disregardthat]

You forgot to square the 2C in going to the second line there.

Thanks, I will correct it.

EDIT: apparently I can't do it now.

 

[lanedance]

yeah I think you have to get in within a certain time frame