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?
 A: 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.
A: I would factor both sides first so that:
$$\sqrt{(x+7)(x+1)}+\sqrt{(x+2)(x+1)}=\sqrt{(6x+13)(x+1)}$$
$$\sqrt{x+1}(\sqrt{x+7}+\sqrt{x+2})=\sqrt{x+1}\sqrt{6x+13}$$
$$\sqrt{x+7}+\sqrt{x+2}=\sqrt{6x+13}$$
$$2x+9+2\sqrt{(x+2)(x+7)}=6x+13$$
$$\sqrt{(x+2)(x+7)}=2x+2$$
$$(x+2)(x+7)=4x^2+8x+4$$
$$3x^2-x-10=0$$
$$(3x+5)(x-2)=0$$
So $x=-\frac{5}{3}$ or $x=2$. However, you find that only $x=2$ works by putting the value into the equation. Do note that while this method may be easier, another solution, $x=-1$ has not been found which would be found by squaring both sides: https://www.symbolab.com/solver/step-by-step/%5Csqrt%7Bx%5E%7B2%7D%2B8x%2B7%7D%2B%5Csqrt%7Bx%5E%7B2%7D%2B3x%2B2%7D%3D%5Csqrt%7B6x%5E%7B2%7D%2B19x%2B13%7D. While this may not be a complete answer, I hope this helped!
A: After writing our equation in the form $$\sqrt{(x+1)(x+7)}+\sqrt{(x+1)(x+2)}=\sqrt{(x+1)(6x+13)}$$ we obtain the domain:
$$(-\infty-7]\cup[-1,+\infty).$$
1. $x\geq-1.$
We see that $-1$ is a root and it remains to solve here 
$$\sqrt{x+7}+\sqrt{x+2}=\sqrt{6x+13}$$ or
$$\sqrt{(x+7)(x+2)}=2(x+1)$$ or
$$3x^2-x-10=0,$$ which gives also $x=2.$


*$x<-1$.


Thus, $x\leq-7$ and we need to solve
$$\sqrt{-x-7}+\sqrt{-x-2}=\sqrt{-6x-13}$$ or
$$\sqrt{(x+7)(x+2)}=-2(x+1),$$ which gives $$3x^2-x-10=0$$ again or
$$(x-2)(3x+5)=0$$ and we see that this equation has no roots for $x\leq-7.$
Id est, we got the answer: $$\{-1,2\}$$
A: When you squared you forget to put the coefficient of 2. 
You should have $2\sqrt{(x^2+8x+7)(x^2+3x+2)}=4(x+1)^2$
Or $\sqrt{(x^2+8x+7)(x^2+3x+2)}=2(x+1)^2$
....
But anyway....
Square again.  But it'll help to factor components.
$(x^2+8x+7)(x^2+3x+2)=4(x+1)^4$
$(x+7)(x+1)(x+2)(x+1)=4(x+1)^4$.
If $x = -1$ we get $0=0$ which could be a solution although we fatored so late in the game we should check if that's extraneous.
Nope... way back in the beginning $x = -1$ would have given us $\sqrt 0 + \sqrt 0 =\sqrt 0$ which is fine.
SO $x =-1$ is a solution.  Divide both sides by $(x+1)^2$ and we tet
$(x+7)(x+2)= 4(x+1)^2$ which is just a quadratic...
$x^2 + 9x + 14 = 4x^2 + 8x + 4$ so
$3x^2 -x -10=0$ which we can solve with the quadratic equation.  (but well have to search for extraneous solutions.)
$x = \frac {1\pm \sqrt{1 + 120}}6 = \frac {1\pm 11}6= 2, -\frac 53$.
But $x =-\frac 53$ give $\sqrt {\frac {25}9 - \frac {40}3+7}= \sqrt{-\frac {32}9}$ which is not real.
So $x=2$ or $x=-1$
