Solve the following equation: $\sqrt{x^2+x}-\sqrt{x^2-x}=x+1$

Question

I have tried many ways but not success

$\sqrt{x^2+x}-\sqrt{x^2-x}=x+1$ $\Leftrightarrow \sqrt{x}(\sqrt{x+1}-\sqrt{x-1})=x+1$

or $\sqrt{x^2+x}-\sqrt{x^2-x}=x+1 \Leftrightarrow \sqrt{\dfrac{x}{x+1}}-\dfrac{\sqrt{x(x-1)}}{x+1}=1 \Leftrightarrow \dfrac{x}{x+1}=1+\dfrac{2\sqrt{x(x-1)}}{x+1}+\dfrac{x(x-1)}{(x+1)^2} \Leftrightarrow \dfrac{(x+1)^2+2\sqrt{x(x-1)}(x+1)+x(x-1)-x(x+1)}{(x+1)^2}=0 \Leftrightarrow x^2+1+2\sqrt{x(x-1)}(x+1)=0 $

I need your help!

Answer 1

Using your first approach, square both sides:

$$x \left((x+1) - 2\sqrt{x^2-1} + (x-1)\right) = (x+1)^2.$$ Now rearrange and collect like terms to isolate the square root:

$$2x \sqrt{x^2 - 1} = 2x^2 - (x+1)^2 = x^2 - 2x - 1.$$

Next, square again:

$$4x^2 (x^2 - 1) = (x^2 - 2x - 1)^2.$$

The resulting quartic has a rather complicated closed form, but it does exist; the only thing left is to eliminate extraneous roots.

Answer 2

After squaring, you face the quartic equation $$3 x^4+4 x^3-6 x^2-4 x-1=0$$

Since $\Delta=-94208$, there are two real roots. Now, forget the nasty formulae and use inspection or graphing; there are two roots "close" to $-2$ and $1.5$. So, use Newton method and obtain the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & -2.00000 \\ 1 & -2.03571 \\ 2 & -2.03396 \end{array} \right)$$

$$\left( \begin{array}{cc} 0 & 1.50000 \\ 1 & 1.32005 \\ 2 & 1.26513 \\ 3 & 1.26015 \\ 4 & 1.26011 \end{array} \right)$$