<s>[INST] 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?
 [/INST] 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
$x=2$
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