1) The product of two consecutive even numbers is 168. Find the numbers.

Let the first even number be \(2x\) and the second even number be \(2x+2\). Then

\begin{align}2x(2x+2) & = 168\\ 4x^2+4x-168 & =0\\ x^2+x-42 & = 0\\ then \; factorising \; gives\\ (x+7)(x-6) & = 0\\ So \; x=-7 \; or \; x & = 6\\ we \; reject \; the \; negative \; solution\\ So \; x & = 6\\ And \; by \; substitution\\ 2(6) & = 12\\ \end{align}

Therefore our first even number is \(12\) and our second is \(14\).

2) The width of a rectangular wall exceeds its height by 2 metres. If the area of the wall is \(143m^2\), find the width of the wall.

Let the width of the wall be \(x\). Then

\begin{align}x(x-2) & = 143\\ x^2-2x-143 & = 0\\ then \; factorising \; gives\\ (x-13)(x+11) & =0\\ So \; x=13 \; or \; x & = -11\\ we \; reject \; the \; negative \; solution\\ So \; x & = 13\\ \end{align}

Therefore the width of the wall is \(13m\)

3) A rectangle has a width of \( (x-1)cm\), a height of \(7cm\), and a diagonal of length \(x \; cm\). Find the width of the rectangle.

\begin{align} By \; Pythagoras\\ (x-1)^2+7^2 & = x^2\\ x^2-2x+1+49 & = x^2\\ -2x+50 & =0\\ 2x & =50\\ x & =25\\ By \; substitution\\ 25-1=24\\ \end{align}

So the width of the rectangle is \(24cm\).

4) One side of a square measures \((x-3)cm\). A rectangle with double the area of the square has a width of \((x+3)cm\) and a height of \((x-4)cm\) . Find two possible values of \(x\).

\begin{align} 2(x-3)^2 & =(x+4)(x-3)\\ 2(x^2-6x+9) & = x^2-x-12\\ 2x^2-12x+18 & = x^2-x-12\\ x^2-11x+30 & = 0\\ then \; factorising \; gives\\ (x-6)(x-5) & =0\\ So \; x= 6 \; or \; x & =5 \end{align}

Therefore the two possible values of \(x\) are \(6\) and \(5\).

5) A man standing on a cliff 40 metres above the sea throws a rock upward. After \(t \; seconds\) the height \(s \) of the rock above the sea is given by the equation \(s= -5t^2+20t+40\). How many seconds before the rock hits the sea?

When the rock hits the sea, the distance \(s\) will be zero. So

\begin{align}-5t^2+10t+40 & = 0\\ -t^2+2t+8 & = 0\\ t^2-2t-8 & = 0\\ (t-4)(t+2) & = 0\\ So \; t = 4 \; or \; t & = -2 \end{align}

We reject the negative solution, as this is a time 2 seconds before the rock was thrown.

Therefore the rock hits the sea after \(4 \; seconds\).

6) A \(1 \;metre \; square\) fountain is surrounded by a path \(x \; metres\) wide. The area of the path is \(24m^2\). Find \(x\).

Each 'corner' of the path has an area of \(x^2 \; m^2\). This leaves four rectangular areas, each of area \(x \; m^2\)

\begin{align} So \; 4x^2+4x & = 24\\ 4x^2+4x-24 & = 0\\ x^2+x-6 & = 0\\ (x+3)(x-2) & = 0 \\ So \; x=2 \; or \; x & =-3\\ we \; reject \; the \; negative \; solution\\ \therefore \; x & =2 \end{align}

Therefore the width of the path is \(2 \; metres\).

7) A cube with side-length \(x \; cm\) has a volume \(19 \; cm^3\) less than a cube of side-length \((x+1)cm\). Find \(x\).

\begin{align}x^3+19 & = (x+1)^3\\ x^3+19 & = (x+1)(x^2+2x+1)\\ x^3+19 & = x^3+3x^2+3x+1\\ 19 & =3x^2+3x+1\\ 0 & = 3x^2+3x-18\\ 0 & = x^2+x-6\\ 0 & = (x+3)(x-2)\\ So \; x=-3 \; or \; x & =2\\ we \; reject \; the \; negative \; solution \\ \therefore \; x & =2 \end{align}

So the value of \(x\) is \(2\).

8) Increasing the speed by \(3km/h\) on a journey of \(40km\) reduces the time taken by \(3 \; hours\). Find the original speed.

We know that \((velocity \; \times \; time) \; = \; distance\)
and also that \((distance \; \div \; velocity) \;= \; time\).

\begin{align}So \quad (x+3)(\frac{40}{x}-3) & =40\\ 40-3x+\frac{120}{x}-9 & = 40\\ -3x^2-9x+120 & =0\\ x^2+3x-40 & =0\\ (x-5)(x+8) & = 0\\ So \; x=5 \; or \; x & =-8 \end{align}

We reject the negative solution
So \(x =5\)

Therefore the original speed was \(5km/h\)