<s>[INST] Solve the following equation: $\sqrt{x^2+x}-\sqrt{x^2-x}=x+1$ 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!
 [/INST] 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
$$\left(
\begin{array}{cc}
n & x_n \\
 0 & -2.00000 \\
 1 & -2.03571 \\
 2 & -2.03396
\end{array}
\right)$$
$$\left(
\begin{array}{cc}
 0 & 1.50000 \\
 1 & 1.32005 \\
 2 & 1.26513 \\
 3 & 1.26015 \\
 4 & 1.26011
\end{array}
\right)$$
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