Find the smallest real value of $x$

Question

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.

Answer 1

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.....

Answer 2

Well, you can clearly see the answer sheet is wrong by plugging in 3 to the original equation, you get $$(3+5)(3^2-3-11)=8(9-3-11)=8(-5)=-40$ on the left hand side, and $8$ on the right hand side, so 3 doesn't satisfy the equation. Ergo, ignore the answer sheet :).