<s>[INST] Where did I go wrong in solving this equation? 
The value of the expression $ax^2 + bx + 1$ are $1$ and $4$ when $x$ takes the values of $2$ and $3$ respectively. Find the value of the expression when $x$ takes the value of $4.$

Here is my first attempt at solving this question:
\begin{align}
2a^2 + 2b + 1 &= 1 \leftarrow\text{(1)} \\
3a^2 + 3b + 1 &= 4 \leftarrow\text{(2)}
\end{align}
$$ \text{From (1):} $$
\begin{align}
2a^2 + 2b + 1 &= 1 \\
2a^2 &= -2b + 1 - 1 \\
2a^2 &= -2b \\
a^2 &= -b \leftarrow\text{(3)}
\end{align}
$$ \text{Substitute (3) into (2):} $$
\begin{align}
3(-b) + 3b + 1 &= 4 \\
-3b + 3b + 1 &= 4 \\
1 &= 4
\end{align}
Here is my second attempt at solving it:
\begin{align}
ax^2 + bx + 1 &= 1 \\
2a^2 + 2b + 1 &= 1 \\
a^2 + b + \frac{1}{2} &= \frac{1}{2} \\
a^2 + b &= \frac{1}{2} - \frac{1}{2} \\
a^2 + b &= 0 \\
4a^2 + 4b &= 0 \\
4a^2 + 4b + 1 &= 1 \leftarrow\text{(1)}
\end{align}
\begin{align}
3a^2 + 3b + 1 &= 4 \\
3a^2 + 3b &= 4 - 1 \\
3a^2 + 3b &= 3 \\
a^2 + b &= 1 \\
4a^2 + 4b &= 4 \\
4a^2 + 4b + 1 &= 5 \leftarrow\text{(2)}
\end{align}
$$ \text{(1) = (2): } 1 = 5 $$
Where did I go wrong?
 [/INST] $$ 4a + 2b + 1 = 1 \leftarrow\text{(1)} $$
$$ 9a + 3b + 1 = 4 \leftarrow\text{(2)} $$
$$ 4a + 2b = 0 $$
$$ 9a + 3b = 3 $$
$$a=1, b=-2$$
</s>