# The Discriminant – A Level Maths Revision

Welcome back there, philomaths! Here, we get serious again with KS5 revision and look at the discriminant in A Level Maths study. Buckle up!

The discriminant (b2 – 4ac) is derived from the quadratic formula and allows you to quickly work out whether a quadratic equation has 2, 1 or 0 real solutions.

If you’re looking for help on how to apply the discriminant to simultaneous equations, check out this page. If you’d like to see the information below in printable or presentable formats, with a larger range of questions, click here. Or, if you want to brush up on the prior knowledge you need for this topic, try these multiple-choice questions. If you’re happy yo understand the discriminant and want to practise, try this colour by number.

You should already be able to use the quadratic formula to find real solutions (where they exist) to each of the following:

1) $x^2 + 10x + 16 = 0$

$x = \frac{-b \, \pm\, \sqrt{b^2\,-\,4ac}}{2a}$

$x = \frac{-10 \, \pm\, \sqrt{10^2\,-\,4\,\times 1\, \times 16}}{2\, \times\, 1}$

$x = \frac{-10 \, \pm\, 6}{2}$

$x = -8 \text{ and } x = -2$

2) $x^2 - 10x + 25 = 0$

$x = \frac{10 \, \pm\, \sqrt{(-10)^2\,-\,4\,\times 1\, \times 25}}{2\, \times\, 1}$

$x = \frac{-10 \, \pm\, 0}{2}$

This time, the number under the square root is 0. The square root of 0 is 0, so you’ll be adding 0 to -10 and subtracting 0 from -10. As both give the same answer, there’s just one solution:

$x = -5$

3) $x^2 + x + 14 = 0$

$x = \frac{-1 \, \pm\, \sqrt{1^2\,-\,4\,\times 1\, \times 14}}{2\, \times\, 1}$

$x = \frac{-1 \, \pm\, \sqrt{-55}}{2}$

This is a problem: $\sqrt{-55}$ does not have a real solution. This means that the equation does not have any real solutions.

These examples show that the value of the expression under the square root, b2 – 4ac, is pivotal in defining whether there will be two real solutions, one real solution or zero real solutions to an equation:

b2 - 4ac > 0 represents two real solutions.
b2 - 4ac = 0 represents one real solution.
b2 - 4ac < 0 represents no real solutions.

b2 – 4ac is called the discriminant because it discriminates between equations with zero, one or two real solutions.

This can also be shown graphically.

The graph of equation 1 above, y = x2 + 10x + 16, with a discriminant of 36, is:

It meets the x-axis twice, indicating that the equation x2 + 10x + 16 = 0 has two real solutions.

The graph of equation 2 above, y = x2 – 10x + 25, with a discriminant of 0, is:

The curve meets the x-axis only once, indicating that the equation x2 – 10x + 25 = 0 has just one real solution.

This is sometimes called a repeated solution, because both brackets in the factorised
equation (x – 5)(x – 5) give the same answer, as do both the positive and negative square roots in the quadratic formula.

The graph of equation 3, y = x2 + x + 14, with a discriminant of -55, is:

Real solutions to the equation x2 + x + 14 = 0 would be where the curve met the x-axis. It doesn’t, hence there are no real solutions.

### Example Question

Given that x2 – 18x + 20 = k has 2 real solutions, find the range of possible values of k.

Having 2 real solutions means that the discriminant is greater than 0.

To find the discriminant, rearrange the quadratic to leave one side as 0:

\begin{aligned} &x^2 - 18x + 20 - k = 0 \\ &a = 1 \\ &b = -18 \\ &c = 20-k \end{aligned}

Substituting into the discriminant gives:

\begin{aligned} b^2 - 4ac &= (-18)^2 - 4 \times 1 \times (20-k) \\ &= 324-4(20-k) \end{aligned}

The discriminant must be greater than 0:

\begin{aligned} &324-4(20-k)>0 \\ &324-80 + 4k > 0 \\ &4k > -244 \\ &k > -61 \end{aligned}

### Practice Questions

1. Find the values of k for which the following equation has just one real solution:
kx2 + x + k = 0

2. y = px2 + 3px – 5 has two real roots; find the range of possible values of p.

3. Find the range of values for p for which the following equation has no real solutions:
3x2 – 10px + 5 + p = -11. Give your answers in surd form.

4. The line with equation y = 7x + g never meets the curve with equation y = gx2 + 4x + 2g.
Find the range of possible values of the constant, g.

5. Sam thinks of two real numbers. When she adds them, the result is 3. When she multiplies them, the result is 100. Show, algebraically, whether she could be correct or has made an error.

1.

\begin{aligned} &b^2 - 4ac = 0 \\ &1^2 - 4 \times k \times k = 0 \\ &4k^2 = 1 \\ & k =\pm 0.5 \end{aligned}

2.

\begin{aligned} &b^2-4ac>0 \\ &(3p)^2 -4\times p \times -5 >0 \\ &9p^2 + 20p > 0 \\&p(9p + 20) > 0 \\ &p< -\frac{20}{9} \text{ or } p>0 \end{aligned}

3.

\begin{aligned} &3x^2 - 10px+16+p = 0 \\ &b^2 - 4ac<0 \\ & (-10p)^2 - 4\times 3 \times (16 + p) <0 \\ & 100p^2 - 192 -12p<0 \\ & 25p^2 - 3p - 48 <0 \end{aligned}

Completing the square gives:

$p<\frac{1}{50}(3+\sqrt{4809}) \text{ or } p>\frac{1}{50}(3-\sqrt{4809})$

4.

\begin{aligned} &7x+g=gx^2+4x+2g \\ &gx^2-3x+g=0\\&b^2-4ac<0 \\ &(-3)^2-4 \times g \times g < 0 \\ &9-4g^2<0 \\ &g>\frac{3}{2},g<-\frac{3}{2} \end{aligned}

5.

Let the starting numbers be x and y.

\begin{aligned} &xy = 100 \\ &x + y = 3 \therefore x = 3-y \\ &(3-y)y = 100 \\ &3y - y^2 - 100 = 0 \\ &a = -1 \\ &b = 3 \\ &c = -100 \\ &b^2- 4ac = 3^2 - 4 \times - 1 \times -100 = -391 \\ &b^2 - 4ac < 0 \end{aligned}

There are no real solutions – Sam calculated incorrectly.

How do you feel now about the discriminant in A Level Maths? For further help and practice opportunities, head over to this resource pack here. But, if you’re ready to keep on rolling on with revision, you can find more of our blogs here! You can also subscribe to Beyond for access to thousands of secondary teaching resources. You can sign up for a free account here and take a look around at our free resources before you subscribe too.