<s>[INST] Solve $\sqrt{x^2+8x+7}+\sqrt{x^2+3x+2}=\sqrt{6x^2+19x+13}$ I tried squaring both sides but doesn't seem like a good idea.
$$x^2+8x+7+\sqrt{(x^2+8x+7)(x^2+3x+2)}+x^2+3x+2=6x^2+19x+13$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4x^2+8x+4$$
$$\sqrt{(x^2+8x+7)(x^2+3x+2)}=4(x+1)^2$$
Is there a better way for solving this equation?
 [/INST] Observe that $(x+1)$ divides all quadratics: the original equation is
$$\sqrt{(x+1)(x+7)} + \sqrt{(x+1)(x+2)} = \sqrt{(x+1)(6x+13)},$$
and rearranging we obtain the following:
$$(\sqrt{x+1})(\sqrt{x+7}+\sqrt{x+2}-\sqrt{6x+13})=0.$$
We are doing some trickery with allowing square roots to venture into $\mathbb C$ here, but note that it is all still correct: for the original equation to have solutions in $\mathbb R$, then either $x \geq -1$ (and all the square roots stay safely within $\mathbb R$) or $x \leq -7$ (in which case all of the square roots are of negative values, so the extra factors of $i$ safely distribute out, assuming we use the principal square root).
The first factor yields a solution of $-1$. The second factor gives solutions when
$$\sqrt{6x+13}= \sqrt{x+7}+\sqrt{x+2},$$
and upon squaring both sides gives
$$6x+13=2x+9+2\sqrt{(x+7)(x+2)},$$
which rearranges to
$$\sqrt{(x+7)(x+2)}=2x+2.$$
Square again, solve the quadratic, and test your solutions to finish.
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