Proving Existence and Uniqueness for x in ℝ in a Quartic Equation

[Harrisonized]

Homework Statement

x/√(x²+y₁²)-(l-x)/√((l-x)²+y₂²)=0

How do I prove that the above equation has a solution for x in ℝ and that the solution is unique?

(y₁, y₂, and l are constants.)

Homework Equations

x√((l-x)²+y₂²)-(l-x)√(x²+y₁²)=0

x√((l-x)²+y₂²)+x√(x²+y₁²)=l√(x²+y₁²)

x[1+√((l-x)²+y₂²)/√(x²+y₁²)]=l=f(x)

I actually don't think it's plausible to isolate x, since it'll result in a quartic equation, which is messy.

The Attempt at a Solution

Existence

Observe that the square root of a real number is always positive. Then lim(x→-∞)f(x)=-∞ and lim(x→∞)f(x)=∞. Since l is a continuous function, there must be at least one value of x that equals the constant l.

Uniqueness

I was told that in general, existence proofs start with the assumption that two different values x₁ and x₂ satisfy the equation, and then show that the two values are actually the same, but I really don't feel like solving a quartic equation.

So I'm going to make an observation. Observe that f(x) is a purely increasing function. Hence, df/dx is always positive.

df/dx=[1+√((l-x)²+y₂²)/√(x²+y₁²)]+x*(d/dx)[√((l-x)²+y₂²)/√(x²+y₁²)]

(d/dx)[√((l-x)²+y₂²)/√(x²+y₁²)]=-(l-x)/[√((l-x)²+y₂²)√(x²+y₁²)]-x√((l-x)²+y₂²)/(x²+y₁²)^3/2

df/dx=[1+√((l-x)²+y₂²)/√(x²+y₁²)]+x*[-(l-x)/[√((l-x)²+y₂²)√(x²+y₁²)]-x√((l-x)²+y₂²)/(x²+y₁²)^3/2]

So messy...

I'm not even sure if df/dx is always positive. It just appears to be that way when I graph it.

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By the way, I know this "proof" isn't rigorous at all, but I've never taken an analysis course. I also don't know how large the difference is between "sufficient" and "complete", but I think the assumption of continuity and limits is sufficient here.

 

[Dick]

You gave up way too easily in trying to isolate x. It's actually really easy. Write the two terms on two different sides of the '=' and square both sides. It falls part pretty easily and the higher degree stuff cancels to give you a quadratic. An EASY quadratic.

 

[Harrisonized]

Actually, the problem I was trying to solve was

x/(c√(x²+y₁²))-(l-x)/(v√((l-x)²+y₂²))=0

I didn't type out the extra coefficients c and v last night because I didn't think it'd make a difference, but apparently it does. Sorry about that. The above definitely does not reduce to an easy quadratic, or I would have done it. :(

Edit: After taking the derivative, I've found that it equals

[1/(c√(x²+y₁²))][1-x²/(x²+y₁²)] + [1/(v√((l-x)²+y₂²))][1-(l-x)²/((l-x)²+y₂²)], which is always positive, since:

x²/(x²+y₁²)

and (l-x)²/((l-x)²+y₂²)

can never be greater than 1.