We know that that the point with \(x\!=\!2\) and \(y\!=\!-\!20\) fits the equation. So we substitute as follows:

The \(x\)-coordinate of the turning point is \(-3\small.\) So \(-3+a=0\) and therefore \(a=3\small.\)

Now we substitute \(x=-1\small,\) \(y=9\) and \(a=3\) into the given equation.

Fact: the graph of \(y=k(x+p)^2+q\) has turning point \((-p,q)\).

In this question, \(p=3\) and \(q=-1\), so the turning point is \((-p,q)=(-3,-1).\)

The axis of symmetry is the vertical line through the turning point, so it is \(x=-3\).

Fact: the graph of \(y=k(x+p)^2+q\) has turning point \((-p,q)\).

The turning point is \((2,5)\) so \(p=-2\) and \(q=5\). Now we know that \(y=k(x-2)^2+5\)

To find \(k\), we substitute the \(x\) and \(y\) coordinates of the other point, \((-1,32)\):

$$ \begin{eqnarray} y &=& k(x-2)^2+5\\[6pt] 32 &=& k(-1-2)^2+5\\[6pt] 32 &=& 9k+5\\[6pt] 9k &=& 27\\[6pt] k &=& 3 \end{eqnarray} $$

The equation of this parabola is therefore \(y=3(x-2)^2+5\)

This has zero on the right hand side and has already been factorised. So it's ready to solve.

\( (2x+1)(3x-10)=0 \) when either \( 2x+1=0 \) or \( 3x-10=0 \)

We get one solution from each linear equation: \( x=-\frac{1}{2} \) or \( x=\frac{10}{3} \).

This simple quadratic equation can be solved by factorising the trinomial. This was only a 2-mark question.

$$ \begin{eqnarray} x^2-11x+24 &=& 0 \\[6pt] (x-3)(x-8) &=& 0 \\[6pt] x-3=0 &\:\ \small\textsf{or}\normalsize\ & x-8=0 \\[6pt] x=3 &\:\ \small\textsf{or}\normalsize\ & x=8 \\[6pt] \end{eqnarray} $$

This quadratic equation arose in context within 2014 Paper 1 Question 13.

The equation needs some rearrangement before factorising.

$$ \begin{eqnarray} 16t-t^2 &=& 60 \\[6pt] -t^2+16t &=& 60 \\[6pt] -t^2+16t-60 &=& 0 \\[6pt] t^2-16t+60 &=& 0 \\[6pt] (t-6)(t-10) &=& 0 \\[6pt] t-6=0 &\:\ \small\textsf{or}\normalsize\ & t-10=0 \\[6pt] t=6 &\:\ \small\textsf{or}\normalsize\ & t=10 \\[6pt] \end{eqnarray} $$

This quadratic equation has a non-unitary coefficient of \(x^2\small.\) This was a 3-mark question.

$$ \begin{eqnarray} 6x^2+13x-5 &=& 0 \\[6pt] (3x-1)(2x+5) &=& 0 \\[6pt] 3x-1=0 &\:\ \small\textsf{or}\normalsize\ & 2x+5=0 \\[6pt] 3x=1 &\:\ \small\textsf{or}\normalsize\ & 2x=-5 \\[6pt] x=\small\frac{1}{3}\normalsize &\:\ \small\textsf{or}\normalsize\ & x=-\small\frac{5}{2}\normalsize \\[6pt] \end{eqnarray} $$

This question requires us to form the equation before solving it.

Let \(x\) represent the number that we are trying to find. So:

$$ \begin{eqnarray} x+x^2 &=& 110 \\[6pt] x^2+x-110 &=& 0 \\[6pt] (x+11)(x-10) &=& 0 \\[6pt] x+11=0 &\:\ \small\textsf{or}\normalsize\ & x-10=0 \\[6pt] x=-11 &\:\ \small\textsf{or}\normalsize\ & x=10 \\[6pt] \end{eqnarray} $$

We are told in the question that \(x\) is negative, so we disregard \(x=10\small.\)

The required number is therefore \(-11\small.\)

Being told to give rounded solutions is a broad clue that \(3x^2-4x-2\) doesn't factorise.

So we use the quadratic formula with \(a=3\small,\) \(b=-4\) and \(c=-2\small.\)

$$ \begin{eqnarray} x &=& \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\[10pt] &=& \frac{4\pm\sqrt{(-4)^2-4(3)(-2)}}{6}\\[10pt] &=& \frac{4\pm\sqrt{16+24}}{6}\\[10pt] &=& \frac{4\pm\sqrt{40}}{6}\\[10pt] &\approx& -0.39\ \textsf{or}\ 1.72 \end{eqnarray} $$

In this question, \(a=1\), \(b=-1\) and \(c=3\)

We calculate the value of the discriminant.

\(b^2-4ac = (-1)^2-4(1)(3) = 1-12 \lt 0\)

A negative discriminant means that \(f(x)\) has no real roots.

In this question, \(a=7\), \(b=5\) and \(c=-1\)

\(b^2-4ac = 5^2-4(7)(-1) = 25+28 \gt 0\)

A positive discriminant means that \(f(x)\) has two distinct real roots.

In this question, \(a=1\small,\) \(b=-6\) and \(c=9\)

\(b^2\!-\!4ac = (-6)^2\!-\!4(1)(9) = 36\!-\!36 = 0\)

A zero discriminant means that \(f(x)\) has two equal real roots. You could also say that \(f(x)\) has one repeated real root.

In this question, \(a=4\small,\) \(b=6\) and \(c=-1\)

\(b^2-4ac = 6^2-4(4)(-1) = 36+16 \gt 0\)

A positive discriminant means that \(f(x)\) has two distinct real roots.

This question combines discriminant and inequalities.

For no real roots, the discriminant must be negative.

We have \(a=p\), \(b=-2\) and \(c=3\small,\) so:

$$ \begin{eqnarray} b^2-4ac &\lt& 0 \\[6pt] (-2)^2-4(p)(3) &\lt& 0 \\[6pt] 4-12p &\lt& 0 \\[6pt] 4 &\lt& 12p \\[6pt] 12p &\gt& 4 \\[6pt] p &\gt& \small\frac{4}{12} \\[6pt] p &\gt& \small\frac{1}{3} \end{eqnarray} $$

The turning point \((5,0)\) lies on the \(x\)-axis.

So the function \(f\) has one real root.

We therefore know that \(b^2-4ac=0\small.\)

Note: A question of this type has never appeared on an SQA paper, and we have our doubts that it ever would, but there's no harm in understanding it!

(a)  We just use the volume formula for a cuboid, and rearrange into the required form.

$$ \begin{gather} 45 = (x+7)(x)(2) \\[6pt] 45 = 2x(x+7) \\[6pt] 45 = 2x^2+14x \\[6pt] 0 = 2x^2+14x-45 \\[6pt] 2x^2+14x-45 = 0 \\[6pt] \end{gather} $$

(b)  Now we solve, using the quadratic formula with \(a=2\small,\) \(b=14\) and \(c=-45\small.\)

$$ \begin{eqnarray} x &=& \frac{-b\pm\sqrt{b^2-4ac}}{2a}\\[10pt] &=& \frac{-14\pm\sqrt{14^2-4(2)(-45)}}{4}\\[10pt] &=& \frac{-14\pm\sqrt{196+360}}{4}\\[10pt] &=& \frac{-14\pm\sqrt{556}}{4}\\[10pt] &\approx& -9.39\ \textsf{or}\ 2.39 \end{eqnarray} $$

As \(x\) is one of the dimensions of the cuboid, we should disregard the negative solution.

So \(x=2.4\) (rounded to 1 d.p.)