Solve the equation over reals: $\sqrt{5x^2 + 27x + 25} - 5\sqrt{x + 1} = \sqrt{x^2 - 4}$.
This problem is adapted from a recent competition. And I can't solve it.
The solutions are $\sqrt{5} + 1$ and $\dfrac{13 + \sqrt{65}}{8}$, as Wolfram Alpha says.
On
Writing $$\sqrt{5x^2+27x+25}=\sqrt{x^2-4}+5\sqrt{x+1}$$ and squaring we get $$4x^2+2x+4=10\sqrt{x+1}\sqrt{x^2-4}$$ dividing by $2$ we get $$2x^2+x+2=5\sqrt{x+1}\sqrt{x^2-4}$$ squaring again we obtain $$4x^4+4x^3+9x^2+4x+4=25(x+1)(x^2-4)$$ expanding and combining like terms $$4x^4-21x^3-16x^2+104x+104=0$$ This can be factorized as $$\left( 4\,{x}^{2}-13\,x-26 \right) \left( {x}^{2}-2\,x-4 \right) =0$$ Such factorization can be found by the following ansatz $$(Ax^2+Bx+C)(ax^2+bx+c)=4x^4-21x^3-16x^2+104x+104$$ and comparing the coefficients.
The domain gives $x\geq2$ and we need to solve $$\sqrt{5x^2+27x+25}=5\sqrt{x+1}+\sqrt{x^2-4}$$ or $$5x^2+27x+25=25(x+1)+10\sqrt{(x+1)(x^2-4)}+x^2-4$$ or $$2x^2+x+2=5\sqrt{(x^2-x-2)(x+2)}$$ or $$\frac{2x^2}{x+2}+1=5\sqrt{\frac{x^2}{x+2}-1}.$$ Now, take $$\sqrt{\frac{x^2}{x+2}-1}=t.$$ Can you end it now?