Solutions of $\sqrt{x^2+5x-14} + |x^2+4x-12|=0$ My attempt:
Given,
$$\sqrt{x^2+5x-14} + |x^2+4x-12|=0 \tag{1}$$
Since $|a|=\sqrt{a^2}$, $$\sqrt{x^2+5x-14}=-\sqrt{(x^2+4x-12)^2}$$
Squaring both sides, $$x^2+5x-14=(x^2+4x-12)^2$$
When I simplify the above, I get two real solutions: $x=2$ and $x=2.138 \text{ (approximately)}$. However, there is only one solution to equation $(1)$ according to Wolfram Alpha, $x=2$.
Is my solution incorrect? If so, where did I go wrong?
 A: Why not factorise first, since both quadratics clearly very easily factorise, into $\sqrt{(x+7)(x-2)} + |(x+6)(x-2)| = 0 \implies$
$\sqrt{(x+7)(x-2)} = -|(x+6)(x-2)| \implies $
$(x+7)(x-2) = ((x+6)(x-2))^2$
This very clearly only has one solution, being $x=2$, since both terms must be zero
A: When you squared both sides, you generated candidate solutions that each had to be checked against the original equation.  As a simpler example:
$\sqrt{4} = x.$
Squaring both sides yields the equation $4 = x^2$, which has the two candidate solutions of $x = \pm 2.$  Each candidate solution must be checked against the original problem to see if it satisfies the problem.
More formally,
you have that if a value satisfies the original equation, and you square both sides, then the value will satisfy the new equation.  However, this is (in general) a one-way implication.  This means that the new equation may have (candidate) solutions that don't satisfy the original equation.
A: When you square, you may get additional solutions. You need to test those solutions in the original equation.
Alternatively,
$\sqrt{x^2+5x-14} + |x^2+4x-12|=0$
$\implies x^2+5x-14 = 0 \ \text { and } x^2+4x-12 = 0$
$\implies (x-2) (x+7) = 0 \ \text { and } (x-2)(x+6) = 0$
So only solution that works is $x = 2$.
A: Both $\sqrt{x^2+5x-14}\geq 0$ and $|x^2+4x-12|\geq 0$
So, both of them must equal to zero, since equation is equal to zero.
$$\sqrt{x^2+5x-14}= 0$$
$$x^2+5x-14=0$$
$x=-7$, or $x=2$
$$|x^2+4x-12|= 0$$
$$x^2+4x-12=0$$
$x=-6$, or $x=2$
Finally
$x=2$
