Solving a radical equation

Question

I have below radical equation which I need to solve for $x$. All other terms, $a,b,c$ are constants. BTW, $x$ is a real positive value.

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

Any help / suggestion to solve this will be greatly appreciated.

Thank you!

Answer 1

I first tried to make the sense of geometry of this problem as described by marwalix and tried to avoid the square root by sine or cosine relations but I failed.

However, I use the Mathematica program which will not give the answer for arbitrary value c. If I assigned c=0.0625, then the answer could be written as follows. $$ \left(\left\{x\to \text{Root}\left[1023.00000000000 \text{$\#$1}^8+4092.00000000000 \text{$\#$1}^7 b+\text{$\#$1}^6 \left(3068.00000000000 a^2+6138.00000000000 b^2\right)+\text{$\#$1}^5 \left(9204.0000000000 a^2 b+4092.00000000000 b^3\right)+\text{$\#$1}^4 \left(3066.00000000000 a^4+10738.0000000000 a^2 b^2+1023.00000000000 b^4\right)+\text{$\#$1}^3 \left(6132.00000000000 a^4 b+6136.00000000000 a^2 b^3\right)+\text{$\#$1}^2 \left(1020.00000000000 a^6-257034.000000000 a^4 b^2+1534.00000000000 a^2 b^4\right)+\text{$\#$1} \left(1020.00000000000 a^6 b-260100.000000000 a^4 b^3\right)-a^8+510.000000000000 a^6 b^2-65025.0000000000 a^4 b^4\&,1\right]\right\}\right) $$ Solve the $\#1$, for specific value of $a$ and $b$, the Root could be solved.

I think there could be better way to avoid this.

Answer 2

The following reduces the equation to a quartic. Let $b=2 a d, x = a(y - d)$, then:

$$\frac{y+d}{\sqrt{(y+d)^2 + 1}} - \frac{y-d}{\sqrt{(y-d)^2 + 1}} = c$$

Squaring:

$$\frac{(y+d)^2}{(y+d)^2 + 1} + \frac{(y-d)^2}{(y-d)^2 + 1} - 2 \;\frac{y^2-d^2}{\sqrt{(y+d)^2 + 1}\sqrt{(y-d)^2 + 1}} = c^2$$

Eliminating denominators:

$$(y+d)^2\big((y-d)^2 + 1\big) + (y-d)^2\big((y+d)^2 + 1\big) -c^2 \big((y+d)^2 + 1\big) \big((y-d)^2 + 1\big) = 2 (y^2-d^2)\sqrt{\big((y+d)^2 + 1\big)\big((y-d)^2 + 1\big)}$$

Collecting and rearranging:

$$(2-c^2)(y^2-d^2)^2 + 2(1-c^2)(y^2+d^2) -c^2 = 2(y^2-d^2)\sqrt{(y^2-d^2)^2 + 2(y^2+d^2)+1}$$

Squaring again gives a quartic in $z = y^2$, which then could technically be solved in radicals.