18 (a) (i) Write $x^2 + 4x - 16$ in the form $(x + a)^2 - b$ - OCR - GCSE Maths - Question 18 - 2018 - Paper 5

To solve for $x$ in the equation $x^2 + 4x - 16 = 0$ using the quadratic formula:

The general form is:
x = rac{-b \\pm \\\sqrt{b^2 - 4ac}}{2a}
where $a = 1$, $b = 4$, and $c = -16$.

1. Calculate the discriminant:
   $b^2 - 4ac = 4^2 - 4 \cdot 1 \cdot (-16) = 16 + 64 = 80$

2. Substitute into the quadratic formula:
   $x = \frac{-4 \pm \sqrt{80}}{2}$
   Simplifying $\\sqrt{80}$ gives us:
   $\sqrt{80} = 4\sqrt{5}$ Therefore:
   $x = \frac{-4 \pm 4\sqrt{5}}{2} = -2 \pm 2\sqrt{5}$

So the solutions are:
$x = -2 + 2\sqrt{5} \quad \text{or} \quad x = -2 - 2\sqrt{5}$.

To sketch the graph of $y = x^2 + 4x - 16$:

1. Identify the turning point from the vertex form we derived, which is at $(-2, -20)$.

   • The vertex is the minimum point since the parabola opens upwards.

2. Calculate additional points by substituting x-values:

   • For $x = 0$:
     $y(0) = 0^2 + 4(0) - 16 = -16$
     Point: $(0, -16)$
   • For $x = -4$:
     $y(-4) = (-4)^2 + 4(-4) - 16 = 0$
     Point: $(-4, 0)$

3. Plot these points and the vertex on the graph. The sketch should show a U-shaped parabola crossing the x-axis at $(-4, 0)$ and $(-2, -20)$ as the turning point.