Suppose that the area bounded by the curve $y = 3x^2 − 12x + 9$, the x-axis and the lines $x = 1$ and $x = k$ is 16. Find $k$. Question: Suppose that the area bounded by the curve $y = 3x^2 − 12x + 9$,
the $x$-axis and the lines $x = 1$ and $x = k$ is $16$. Find $k$.
I tried but found no solution.
What I did: 
$$\int_1^k (3x^2-12x+9) \,dx = \big[x^3-6x+9x \big]_1^k = k^3-6k^2+9k-4 =16$$ 
$$\Rightarrow k^3-6k^2+9k-20 = 0$$ 
Then after simplifying I got: 
$$(k-5)(k^2-k+4) = 0$$
This means $k = 5$, because $(k^2-k+4)$ has no real roots, but this is wrong. The reason is after drawing the graph, the area under the curve, from $x=1$ to $x=5$ is:
$$-\int_1^3 (3x^2-12x+9) \, dx + \int_3^5 (3x^2-12x+9) \, dx = 24 \ne 16$$ 
Can anyone help? What mistake did I make?

As I set the question, but do not have the reputation to comment, I will say what the question should have been, and answer it. I will also provide an answer to the question you attempted, and point out your flaw.
There was a typo on the sheet: the 1 should have been a -1 (this was corrected at some point though).
Therefore, the question goes as follows:
$$
\begin{align*}
16&=\int\limits_{-1}^k3x^2-12x+9\mathrm{d}x &\text{NOTE: Do not omit the dx (as in the OP). You'll drop a mark.}\\
&=\left[x^3-6x^2+9x\right]_{-1}^k\\
&=k^3-6k^2+9k-(-1-6-9)\\
\Rightarrow 0&=k^3-6k^2+9k\\
\Rightarrow k&=0\text{ or }k=3
\end{align*}
$$
We discard the $k=3$ solution so the answer is $k=0$. I will leave you to work out why we discard $k=3$ rather than $k=0$...

If we take the lower limit to be $x=1$, then note that the area between $x=1$ and $x=3$ (the root we care about) is $4$. Hence, $k>3$. Therefore, the equation you should have been solving is the following.
$$
\begin{align*}
16&=-\int\limits_{-1}^33x^2-12x+9\mathrm{d}x+\int\limits_{3}^k3x^2-12x+9\mathrm{d}x\\
&=4+\int\limits_{3}^k3x^2-12x+9\mathrm{d}x
\end{align*}
$$
This gives you the equation $(k-3)^2k=12$. The only real root is $k\approx4.6129$.
