Express $y = 4x - 4x^2 - 3$ in the form $p(x + q)^2 + r$ where $p$, $q$ and $r$ are constants. Hence, explain whether $4x - 4x^2 - 3 = 0$ has any real solutions.

1. Rearrange the terms and factor out the coefficient of $x^2$ from the $x$ terms.

$y = -4x^2 + 4x - 3$
$y = -4(x^2 - x) - 3$

2. Complete the square for the expression inside the bracket.

$y = -4\left[ \left(x - \frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 \right] - 3$
$y = -4\left[ \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} \right] - 3$

3. Expand and simplify to reach the final form.

$y = -4\left(x - \frac{1}{2}\right)^2 + 1 - 3$
$y = -4\left(x - \frac{1}{2}\right)^2 - 2$

1. Express the quadratic equation using the completed square form from part (a)

From part (a), $4x - 4x^2 - 3 = -4\left(x - \frac{1}{2}\right)^2 - 2$.
So the equation becomes $-4\left(x - \frac{1}{2}\right)^2 - 2 = 0$.

2. Explain why there are no real solutions based on the maximum value

For any real $x$, $\left(x - \frac{1}{2}\right)^2 \geqslant 0$, which means $-4\left(x - \frac{1}{2}\right)^2 \leqslant 0$.
Therefore, $-4\left(x - \frac{1}{2}\right)^2 - 2 \leqslant -2$.
Since the maximum value of the expression is $-2$, it can never be equal to $0$. Hence, the equation has no real solutions.