6 (a) Write $\sqrt{80}$ in the form $c\sqrt{5}$, where $c$ is a positive constant - Edexcel - A-Level Maths Pure - Question 8 - 2014 - Paper 1

Given the length of rectangle R is $(1 + \sqrt{5})$ cm and the area is $\sqrt{80}$ cm$^2$, we can set up the equation for the width $w$:

$$\text{Area} = \text{Length} \times \text{Width} \implies \sqrt{80} = (1 + \sqrt{5}) \times w.$$

Now, we need to isolate $w$ by dividing both sides:

$$w = \frac{\sqrt{80}}{1 + \sqrt{5}}.$$

To simplify this expression, we can multiply the numerator and denominator by the conjugate of the denominator, $1 - \sqrt{5}$:

$$\begin{align*} w & = \frac{\sqrt{80}(1 - \sqrt{5})}{(1 + \sqrt{5})(1 - \sqrt{5})} \\ & = \frac{\sqrt{80}(1 - \sqrt{5})}{1 - 5} \\ & = \frac{\sqrt{80}(1 - \sqrt{5})}{-4}. \end{align*}$$

Now, we can simplify $\sqrt{80}$:

$\sqrt{80} = 4\sqrt{5}.$

Thus, we substitute it back:

w = \frac{4\sqrt{5}(1 - \sqrt{5})}{-4} = -\sqrt{5}(1 - \sqrt{5}) = (\sqrt{5} - 5).$$ Hence, the width $w$ can be expressed as: $$w = -5 + 1\sqrt{5},$$ where $p = -5$ and $q = 1$.