National 5 Maths
Quadratics

Questions like this check your basic understanding of what an equation is.

An equation is just a rule relating \(x\)-coordinates (\(x\small,\) for short) and \(y\)-coordinates (\(y\small,\) for short).

In this example, we know that that the point with \(x\!=\!2\) and \(y\!=\!-20\) fits the equation \(y=kx^2\small.\)

So we substitute as follows:

Note: Questions like this are only worth 2 marks. For example: 2014 P1 Q7 and 2021 P1 Q6 .

Video solution by Clelland Maths 

You need to know that the graph of \(y=(x+a)^2+b\) has a vertical line of symmetry \(x=-a\small.\)

In this example, we have been told that the axis of symmetry is \(x=-3\small.\) So \(-a=-3\small,\) giving \(a=3\small.\)

We have also been told that the point with \(x=-1\) and \(y=9\) satisfies the given equation.

So we substitute as follows:

Note: Questions of this type are typically worth 3 marks. See, for example: 2017 P1 Q14 .

Video solution by Clelland Maths 

(a)  To answer this, you need to have learned that the graph of \(y=k(x+p)^2+q\) has turning point \((-p,\,q)\small.\)

In this question, \(p=3\) and \(q=-1\small.\)

So this turning point is \((-p,\,q)=(-3,\,-1)\small.\)

(b)  You also need to have learned that the the graph of \(y=k(x+p)^2+q\) has axis of symmetry \(x=-p\small.\)

Another way of remembering this is that the axis of symmetry is the vertical line through the turning point.

So, in this example, the axis of symmetry is \(x=-3\small.\)

Video solution by Clelland Maths 

This question gives us no hint about the form of the equation.

To get started, we need to use the fact that the graph of \(y=k(x+p)^2+q\) has turning point \((-p,\,q)\small.\)

The turning point of the given graph is \((2,\,5)\) so \(p=-2\) and \(q=5\small.\)

At this point in the working, we know that that \(y=k(x-2)^2+5\) but we do not know \(k\) yet.

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

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

So the equation of this parabola is \(y=3(x-2)^2+5\small.\)

Note: You could expand this and answer in the form \(y=3x^2-12x+17\) if you prefer, but it isn't necessary.

Video solution by Clelland Maths 

This equation has \(0\) on the right hand side. The left hand side has already been factorised. So it's ready to solve.

It is important to understand that \((2x+1)(3x-10)=0\) when either \(2x+1=0\) or \(3x-10=0\small.\)

So the full solution looks like this:

$$ \begin{gather} (2x+1)(3x-10)=0\\[7pt] 2x+1=0\:\:\small\textsf{or}\normalsize\:\:3x-10=0\\[7pt] 2x=-1\:\:\small\textsf{or}\normalsize\:\:3x=10\\[6pt] x=\small-\frac{1}{2}\normalsize\:\:\small\textsf{or}\normalsize\:\:x=\small\frac{10}{3}\\[6pt] \end{gather} $$

Note: These solutions should be left as fractions.

Video solution by Clelland Maths 

This simple 2-mark quadratic equation can be solved by factorising the trinomial:

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

Video solution by Clelland Maths 

This quadratic equation arose in context within 2014 P1 Q13 .

The equation needs some rearrangement before factorising. We always need zero on one side.

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

Note: We multiplied the whole equation through by \(-1\) between the third and fourth lines above. That's a very common trick to give us a positive coefficient of the squared term, so that it's easier to factorise.

Video solution by Clelland Maths 

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

$$ \begin{gather} 6x^2+13x-5=0 \\[7pt] (3x-1)(2x+5)=0 \\[7pt] 3x-1=0\:\:\small\textsf{or}\normalsize\:\:2x+5=0 \\[7pt] 3x=1\:\:\small\textsf{or}\normalsize\:\:2x=-5 \\[6pt] x=\small\frac{1}{3}\normalsize\:\:\small\textsf{or}\normalsize\:\:x=\small-\frac{5}{2}\normalsize \\[6pt] \end{gather} $$

Video solution by Clelland Maths 

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

Let \(x\) represent the number that we are trying to find. So \(x+x^2=110\small.\)

Now we solve by rearranging and factorising:

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

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

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

Video solution by Clelland Maths 

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 \(c=-2\small.\)

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

Video solution by Clelland Maths 

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

We calculate the value of the discriminant like this:

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

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

Video solution by Clelland Maths 

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

We calculate the value of the discriminant like this:

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

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

Video solution by Clelland Maths 

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

We calculate the value of the discriminant like this:

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

The discriminant is zero, so \(f(x)\) has two equal real roots. Another way of saying this is that \(f(x)\) has one repeated real root. Answering using either wording will receive full marks.

Video solution by Clelland Maths 

In this 2-mark question, \(a=4\small,\) \(b=6\) and \(c=-1\small.\)

We calculate the value of the discriminant like this:

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

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

Video solution by Clelland Maths 

This question combines discriminant and inequalities.

For no real roots, the discriminant must be negative.

From the equation, \(a=p\small,\,\) \(b=-2\) and \(c=3\small,\) so:

$$ \begin{gather} b^2-4ac\lt 0 \\[8pt] (-2)^2-4(p)(3)\lt 0 \\[8pt] 4-12p\lt 0 \\[8pt] 4\lt 12p \\[6pt] 12p\gt 4 \\[6pt] p\gt \small\frac{4}{12} \\[6pt] p\gt \small\frac{1}{3} \end{gather} $$

Video solution by Clelland Maths 

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: This hasn't appeared in an exam, and perhaps it never will, but it's still worth understanding!

Video solution by Clelland Maths 

(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.)

Video solution by Clelland Maths 

Being told to give rounded solutions is a broad clue that \(3x^2+8x+1\) doesn't factorise.

So we use the quadratic formula with \(a=3\small,\,\) \(b=8\) and \(c=1\small.\)

$$ \begin{eqnarray} x &=& \frac{-b\pm\sqrt{b^2-4ac\,}}{2a}\\[10pt] &=& \frac{-8\pm\sqrt{8^2-4\times 3\times 1\,}}{2\times 3}\\[10pt] &=& \frac{4\pm\sqrt{64-12\,}}{6}\\[10pt] &=& \frac{4\pm\sqrt{52\,}}{6}\\[10pt] &\approx& -2.54\ \textsf{or}\ -0.13 \end{eqnarray} $$

Video solution by Clelland Maths