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 $L = 1 + \sqrt{5}$ cm and the area $A = \sqrt{80}$ cm$^2$, we can establish the relationship:

$A = L \cdot W$

where $W$ is the width. Thus,

$\sqrt{80} = (1 + \sqrt{5})W$

We already determined that $\sqrt{80} = 4\sqrt{5}$, so substituting this in gives us:

$4\sqrt{5} = (1 + \sqrt{5})W$

Next, we solve for $W$ by dividing both sides by $(1 + \sqrt{5})$:

$W = \frac{4\sqrt{5}}{1 + \sqrt{5}}$

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

$W = \frac{4\sqrt{5}(1 - \sqrt{5})}{(1 + \sqrt{5})(1 - \sqrt{5})}$

Calculating the denominator:

$(1 + \sqrt{5})(1 - \sqrt{5}) = 1^2 - (\sqrt{5})^2 = 1 - 5 = -4$

Thus,

$W = \frac{4\sqrt{5}(1 - \sqrt{5})}{-4} = -\sqrt{5}(1 - \sqrt{5}) = -\sqrt{5} + 5$

Rearranging gives:

$W = 5 - \sqrt{5}$

Now, expressing this in the form $p + q\sqrt{5}$, we identify:

$p = 5, \; q = -1$