Learn9709Paper 1 · Pure Mathematics 1QuadraticsThe discriminant and the nature of roots

Quadratics · Subtopic 3 of 5

The Discriminant and the Nature of Roots

How $b^2-4ac$ tells you whether a quadratic has two, one or no real roots without solving it, how to find unknown constants from that, and how to show an expression is always positive.

Cambridge 9709 Paper 1Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2AQA 7357 Paper 3≈ 40 minRevision 6 · updated 2026-10-27

Section 1 of 14

Learning objectives

Calculate the discriminant $b^2-4ac$ and use it to state the number of real roots.
Choose the correct condition: $>0$ , $=0$ , $\ge0$ or $<0$ , from the wording of a question.
Find an unknown constant, or the range of values of a constant, from a condition on the roots.
Show that a quadratic expression is positive (or negative) for all real $x$ .

Pause and think

In the quadratic formula $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ , what happens to the " $\pm$ " part if $b^2-4ac=0$ ? What if $b^2-4ac$ is negative?

Section 2 of 14

Core idea

Start hereCan the ball reach 25 metres?

The ball from earlier has height $h=1+20t-5t^2$ . Can it ever reach $25$ m? Set $1+20t-5t^2=25$ , i.e. $5t^2-20t+24=0$ . You could try to solve it, but there's a faster question: does it have any solutions at all? One number, worked out in a few seconds, answers that. It's called the discriminant because it discriminates (tells apart) the three possible situations.

The big idea

For $ax^2+bx+c=0$ , the discriminant $\Delta=b^2-4ac$ decides how many real roots there are: $\Delta>0$ two distinct roots; $\Delta=0$ one repeated root; $\Delta<0$ no real roots. Graphically: the curve crosses the $x$ -axis twice, touches it once, or misses it.

1. Where the discriminant comes from

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

$\Delta=b^2-4ac$: the expression under the square root. ( $\Delta$ is the Greek capital "delta".)

Everything in the formula is an ordinary number except the square root. A square root behaves in three different ways depending on what's inside:

Three cases

$\Delta>0:\quad x_1=\frac{-b+\sqrt\Delta}{2a}\neq x_2=\frac{-b-\sqrt\Delta}{2a}$
$\sqrt\Delta$ is a positive number, so adding and subtracting it give two different answers.
$\Delta=0:\quad x=\frac{-b\pm0}{2a}=-\frac b{2a}$
$\sqrt0=0$ , so $+$ and $-$ give the same answer: one repeated root.
$\Delta<0:\quad\sqrt\Delta\text{ is not a real number}$
No real number squared is negative, so there are no real roots.

\Delta>0

: 2 real roots.

\;\Delta=0

: 1 (repeated) real root.

\;\Delta<0

: no real roots.

y = x² − 4x + c: discriminant = 16 − 4c

c = 1.00

For

c<4

the discriminant

16-4c

is positive and there are two crossing points. At

c=4

it is zero and the curve touches the axis. For

c>4

it is negative and the curve misses the axis.

Edge case"Real roots" means $\Delta\ge0$ If a question says "has real roots" without "distinct", it includes the repeated case: use

\Delta\ge0

. "Two distinct real roots" means

\Delta>0

2. Back to the ball

Worked exampleDoes $5t^2-20t+24=0$ have solutions?

Decide whether the ball with $h=1+20t-5t^2$ can reach $25$ m. What's the greatest height it can reach?

Goal: Count real roots of $5t^2-20t+24=0$ ; then find the height where $\Delta=0$ .
Strategy: Compute $\Delta$ . For the second part, replace $25$ by an unknown height $H$ and make $\Delta=0$ .

$\Delta=(-20)^2-4(5)(24)=400-480=-80$
$a=5$ , $b=-20$ , $c=24$ .
$\Delta<0$ : no real $t$ . The ball never reaches $25$ m.
$1+20t-5t^2=H\;\Longrightarrow\;5t^2-20t+(H-1)=0$
General height $H$ ; rearrange to $=0$ .
$\Delta=400-20(H-1)=0\;\Longrightarrow\;H=21$
The highest reachable height is reached exactly once: repeated root, $\Delta=0$ .

AnswerNo; the maximum height is

21

✓ CheckMatches the completing-the-square answer:

h=-5(t-2)^2+21

✓.

3. Unknown coefficients

Worked exampleEqual roots

Find $k$ such that $x^2+kx+9=0$ has equal roots, and find the root.

Goal: Value(s) of $k$ , then the root.
Strategy: Equal roots ⇔ $\Delta=0$ .

$k^2-4(1)(9)=0$
$a=1$ , $b=k$ , $c=9$ .
$k^2=36\Rightarrow k=\pm6$
Both signs.
$k=6:\ x^2+6x+9=(x+3)^2=0\Rightarrow x=-3$
A perfect square, as expected.
$k=-6:\ (x-3)^2=0\Rightarrow x=3$
Likewise.

Answer

k=6

(root

-3

) or

k=-6

(root

3

✓ CheckRepeated root

=-\frac b{2a}

-\frac62=-3

✓.

Worked exampleTwo distinct real roots: an inequality for $k$

Find the values of $k$ for which $kx^2+4x+(k-3)=0$ has two distinct real roots.

Goal: A range of $k$ .
Strategy: Set $\Delta>0$ ; it becomes a quadratic inequality in $k$ .

$16-4k(k-3)>0$
$a=k$ , $b=4$ , $c=k-3$ .
$16-4k^2+12k>0$
Expand.
$k^2-3k-4<0$
Divide by $-4$ : the inequality flips.
$(k-4)(k+1)<0\Rightarrow-1<k<4$
A ∪-shaped parabola is negative between its roots.
Exclude $k=0$ : then the equation isn't quadratic ( $4x-3=0$ has only one root).

Answer

-1<k<4

k\neq0

✓ CheckTry

k=1

x^2+4x-2=0

\Delta=16+8=24>0

✓. Try

k=5

\Delta=16-40<0

✓ (correctly excluded).

Common misconceptionForgetting a ≠ 0 When the

x^2

coefficient contains

k

, check whether that value of

k

destroys the quadratic.

4. Always positive: proving it

a>0

and

\Delta<0

, then

ax^2+bx+c>0

for all

x

$ax^2+bx+c=a\left(x+\tfrac b{2a}\right)^2+\tfrac{4ac-b^2}{4a}$
Completed-square form.
$\tfrac{4ac-b^2}{4a}=\tfrac{-\Delta}{4a}>0$
$\Delta<0$ makes $-\Delta>0$ , and $a>0$ .
$a\left(x+\tfrac b{2a}\right)^2\ge0$
$a>0$ times a square.
$ax^2+bx+c>0$
Something $\ge0$ plus something $>0$ .

The graph lies entirely above the

x

-axis. (If

a<0

and

\Delta<0

, it's always negative.)

Pause and think

Show $2x^2-3x+5$ is positive for all $x$ .

5. Does a line meet a curve?

Substituting a line into a curve gives a quadratic; its discriminant counts the meeting points. $\Delta=0$ means the line is a tangent.

Worked exampleTangent to $y=x^2+1$

Find $m$ so that $y=mx$ is a tangent to $y=x^2+1$ .

Goal: Values of $m$ giving exactly one meeting point.
Strategy: Equate, rearrange, set $\Delta=0$ .

$x^2+1=mx\;\Rightarrow\;x^2-mx+1=0$
Same $y$ at a meeting point.
$m^2-4=0\;\Rightarrow\;m=\pm2$
Tangent ⇔ one repeated root.

Answer

m=2

m=-2

✓ Check

m=2

x^2-2x+1=(x-1)^2

, touching at

(1,2)

✓.

Reading the discriminant

Drawing diagram…

For a line meeting a curve, the same three outcomes mean: cuts twice, tangent, misses.

Summary

$\Delta=b^2-4ac$ is the part under the square root.
$>0$ , $=0$ , $<0$ : two, one, no real roots.
"Real roots": $\Delta\ge0$ .
$a>0$ and $\Delta<0$ ⇒ always positive.
Line meets curve: $\Delta=0$ ⇔ tangent.

Discriminant: $b^2-4ac$ ; decides the number of real roots.
Repeated root: Both roots equal; the graph touches the axis.
Tangent: A line touching a curve at one point.

Three parabolas, three discriminants

The discriminant

b^2-4ac

tells you how many times the graph meets the

x

-axis: twice, once (touching) or not at all.

HistoryWho named it? The English mathematician James Joseph Sylvester introduced the word "discriminant" in 1851. He invented a lot of mathematical vocabulary and once called himself the "mathematical Adam" because he gave names to so many things.

In the real worldVideo games use the discriminant millions of times a second. To draw realistic light, a computer checks whether each ray of light hits an object. For a sphere, that check is a quadratic equation, and the sign of its discriminant says "hit", "just grazes" or "miss", before any square root is taken.

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