How to solve this kind of equation?

Question

I have an equation (in my homework) of the form

$a=\sqrt{x^2 + b^2} + \sqrt{x^2 + c^2}$

which I would like to solve for $x$. I am not sure how best to proceed. My thought is to square both sides of the equation, which gives me

$a^2=b^2+c^2+2x^2+\sqrt{(b^2+x^2)\times(c^2+x^2)}$

and then

$a^2=b^2+c^2+2x^2+\sqrt{b^2c^2 + b^2x^2 + c^2x^2 + x^4}$

but I am not sure what to do with the square root in order to eventually solve for $x$. How do I finish solving this, or am going the wrong way?

Answer 1

Suppose that our equation $$\sqrt{x^2+b^2}+\sqrt{x^2+c^2}=a.\tag{1}$$ holds. Assume $a\ne 0$ and $b\ne c$. Flip both sides over. We get $$\frac{1}{\sqrt{x^2+b^2}+\sqrt{x^2+c^2}}=\frac{1}{a}.$$ Multiply top and bottom on the left by $\sqrt{x^2+b^2}-\sqrt{x^2+c^2}$. We get $$\frac{\sqrt{x^2+b^2}-\sqrt{x^2+c^2}}{b^2-c^2}=\frac{1}{a}.$$ Conclude that $$\sqrt{x^2+b^2}-\sqrt{x^2+c^2}=\frac{b^2-c^2}{a}.\tag{2}$$

Add (1) and (2). Now we know $2\sqrt{x^2+b^2}$, and the rest is downhill.

Answer 2

Isolate $\sqrt{x^2+c^2}$, then square both sides. Then isolate the last radical standing. I trust you can handle it from there.