<s>[INST] Find the smallest real value of $x$ I don't know why my answer is different from the answer sheet. 

Find the smallest real value of $x$ that satisfies the equation: $(x+5)(x^2-x-11)=x+5$

Here is what I did. This equation can be rewritten as $(x+5)(x^2-x-12)=(x+5)(x-4)(x+3)=0$, which give $$(x+5)=0$$ or $$(x-4)=0$$ or $$(x+3)=0$$ Then the smallest real number I got is -5. But the answer sheet says 3.
 [/INST] If $x \neq -5$ then divide by $x+5$ and have
$$x^2 - x - 11 = 1$$
$$x^2 - x - 12 =0$$
Solve it:
$$x = \frac{1\pm\sqrt{1 + 48}}{2}$$
Hence you get two solutions:
$$x_1 = +4 ~~~~~~~ x_2 = -3$$
So.....
In any case
$x = -5$ solves the equation too.
So.....
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