Mathematics 9709 · Pure Mathematics 1

The discriminant

One number, $b^2-4ac$ (the bit under the square root), tells you how many real roots a quadratic has.
It works without solving the equation.

Where it comes from

In the quadratic formula everything is fixed except the square root. Only its sign matters here. So we give it a name: $\Delta = b^2-4ac$ .

$\Delta = b^2-4ac$

You cannot take the square root of a negative and get a real answer. So whether $b^2-4ac$ is positive, zero, or negative decides everything, before you solve anything.

Why the formula leads here

Look at the quadratic formula $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ . Once you know $a$ , $b$ , and $c$ , every part is fixed except $\sqrt{b^2-4ac}$ . The $\pm$ usually gives two answers. What changes that number is the value under the root. That is why this one number, the discriminant $\Delta$ , matters.

The rule

Rearrange to $=0$ , read off $a,b,c$ , compute $b^2-4ac$ , and read its sign.

Rearrange to $=0$ : move every term to one side so the other is zero (e.g. $x^2+5x=6$ becomes $x^2+5x-6=0$ ).
Read off $a$ , $b$ , $c$ from that $=0$ form (so here $c=-6$ , not $6$ ).
Compute $b^2-4ac$ .
Read its sign off the table below.

Discriminant	Real roots	Curve & x-axis
$b^2-4ac>0$	two distinct	crosses twice
$b^2-4ac=0$	one repeated	touches (tangent)
$b^2-4ac<0$	no real roots	never meets

The calculator does the sum. The skill is reading off $a,b,c$ with the right signs (and remembering $b^2$ is never negative).

See the rule in three quick examples

Positive → $x^2-5x+6=0$ , $a=1,\ b=-5,\ c=6$ : $(-5)^2-4(1)(6)=25-24=1>0$ . It factorises as $(x-2)(x-3)=0$ → $x=2$ or $x=3$ , two roots.

Zero → $x^2-6x+9=0$ , $a=1,\ b=-6,\ c=9$ : $(-6)^2-4(1)(9)=36-36=0$ . It is $(x-3)(x-3)=0$ , and $\pm\sqrt{0}$ adds nothing, so the two roots become the same number → one repeated root $x=3$ .

Negative → $x^2+2x+5=0$ , $a=1,\ b=2,\ c=5$ : $(2)^2-4(1)(5)=4-20=-16<0$ . The formula needs $\sqrt{-16}$ , which is not a real number → no real roots.

Your turn— tap to reveal the worked answer (quick check)

How many real roots does $x^2+x-6=0$ have?

$b^2-4ac=(1)^2-4(1)(-6)=1+24=25$ , which is positive.

Answer

\text{two real roots}

See it move

As the parabola moves, two roots come together into one (a tangent), then disappear. This happens exactly as $b^2-4ac$ passes through zero.

The dots are where it crosses the $x$ -axis. These are the real roots. Drag the slider and watch them move.

b² − 4ac = 12.0 → two real roots

What the exam asks

The exam hides the discriminant inside geometry words. Turn words about a line and a curve into a $b^2-4ac$ condition.

Wording	Condition	Seen in
is a tangent to	$b^2-4ac=0$	(9709/13 Jun 2021 Q3)
meets at two distinct points	$b^2-4ac>0$	(9709/12 Mar 2022 Q2)
no real roots / does not meet	$b^2-4ac<0$	(9709/12 Jun 2023 Q3; 9709/11 Nov 2020 Q1)

The key step: when a line meets a curve, put the line into the curve and rearrange to $=0$ first. Then test the discriminant of that quadratic, not the curve's.

Worked examples

“Equal roots” and “tangent” both mean $b^2-4ac=0$ . Set it up and solve.

Worked example

Standard case — find p so that $x^2+px+9=0$ has equal roots

“Equal roots” means one repeated root, so need $b^2-4ac=0$ , with $a=1,\ b=p,\ c=9$ .
Write the condition: $p^2-4(1)(9)=0$ .
Tidy and rearrange: $p^2-36=0\Rightarrow p^2=36$ .
Square-root both ways:

Answer

p=6\ \text{or}\ p=-6

Worked example

Exam level — find k so that $y=2x+k$ is a tangent to $y=x^2-4x+1$

Tangent → one repeated root → $b^2-4ac=0$ . Set line = curve: $x^2-4x+1 = 2x+k$ .
Bring everything across to $=0$ : $x^2-6x+(1-k)=0$ , so $a=1,\ b=-6,\ c=1-k$ .
Set discriminant to zero: $(-6)^2-4(1)(1-k)=0$ .
Expand and solve: $36-4+4k=0\Rightarrow 32+4k=0$ .

Answer

k=-8

Show the full working

$x^2-4x+1 = 2x+k$

$x^2-4x-2x+1-k=0$

Collect the $x$ terms:

$x^2-6x+(1-k)=0$

Set the discriminant to zero, minding the signs (since $-4\times(-k)=+4k$ ):

$(-6)^2-4(1)(1-k)=0$

$36-4(1-k)=0\quad\Rightarrow\quad 36-4+4k=0$

$32+4k=0$

Notice we never found where the line touches. The discriminant answered the question on its own. This is the exact step in 9709/13 Jun 2021 Q3.

Where the marks go

Common mistake

A tangent is one repeated root, so

b^2-4ac=0

, not

>0

. And “no real roots” needs a strict

b^2-4ac<0

, not

\le

. The equals case is a repeated root, and that does count as meeting.

Common mistake

Always rearrange to

=0

before reading off

a,b,c

, and mind the sign of

b

when you square it: with

b=-6

b^2=(-6)^2=36

, not

-36

Now you try

Find the set of values of $m$ for which the line $y=mx-3$ does not meet the curve $y=2x^2+5$ . (9709/11 Nov 2020 Q1)

Your turn— tap to reveal the worked answer (9709/11 Nov 2020 Q1)

“Does not meet” → no real roots → $b^2-4ac<0$ .
Set line = curve and rearrange to $=0$ : $2x^2-mx+8=0$ , so $a=2,\ b=-m,\ c=8$ .
Apply the condition: $(-m)^2-4(2)(8)<0 \Rightarrow m^2-64<0$ .
Quadratic inequality; roots $\pm 8$ , negative between them.

Answer

-8<m<8

PreviousThe quadratic graph NextQuadratic inequalities

The discriminant

One number, $b^2-4ac$ (the bit under the square root), tells you how many real roots a quadratic has.
It works without solving the equation.

Where it comes from

In the quadratic formula everything is fixed except the square root. Only its sign matters here. So we give it a name: $\Delta = b^2-4ac$ .

$\Delta = b^2-4ac$

You cannot take the square root of a negative and get a real answer. So whether $b^2-4ac$ is positive, zero, or negative decides everything, before you solve anything.

Why the formula leads here

The rule

Rearrange to $=0$ , read off $a,b,c$ , compute $b^2-4ac$ , and read its sign.

Rearrange to $=0$ : move every term to one side so the other is zero (e.g. $x^2+5x=6$ becomes $x^2+5x-6=0$ ).
Read off $a$ , $b$ , $c$ from that $=0$ form (so here $c=-6$ , not $6$ ).
Compute $b^2-4ac$ .
Read its sign off the table below.

Discriminant	Real roots	Curve & x-axis
$b^2-4ac>0$	two distinct	crosses twice
$b^2-4ac=0$	one repeated	touches (tangent)
$b^2-4ac<0$	no real roots	never meets

The calculator does the sum. The skill is reading off $a,b,c$ with the right signs (and remembering $b^2$ is never negative).

See the rule in three quick examples

Positive → $x^2-5x+6=0$ , $a=1,\ b=-5,\ c=6$ : $(-5)^2-4(1)(6)=25-24=1>0$ . It factorises as $(x-2)(x-3)=0$ → $x=2$ or $x=3$ , two roots.

Negative → $x^2+2x+5=0$ , $a=1,\ b=2,\ c=5$ : $(2)^2-4(1)(5)=4-20=-16<0$ . The formula needs $\sqrt{-16}$ , which is not a real number → no real roots.

Your turn— tap to reveal the worked answer (quick check)

How many real roots does $x^2+x-6=0$ have?

$b^2-4ac=(1)^2-4(1)(-6)=1+24=25$ , which is positive.

Answer

\text{two real roots}

See it move

As the parabola moves, two roots come together into one (a tangent), then disappear. This happens exactly as $b^2-4ac$ passes through zero.

The dots are where it crosses the $x$ -axis. These are the real roots. Drag the slider and watch them move.

b² − 4ac = 12.0 → two real roots

What the exam asks

The exam hides the discriminant inside geometry words. Turn words about a line and a curve into a $b^2-4ac$ condition.

Wording	Condition	Seen in
is a tangent to	$b^2-4ac=0$	(9709/13 Jun 2021 Q3)
meets at two distinct points	$b^2-4ac>0$	(9709/12 Mar 2022 Q2)
no real roots / does not meet	$b^2-4ac<0$	(9709/12 Jun 2023 Q3; 9709/11 Nov 2020 Q1)

The key step: when a line meets a curve, put the line into the curve and rearrange to $=0$ first. Then test the discriminant of that quadratic, not the curve's.

Worked examples

“Equal roots” and “tangent” both mean $b^2-4ac=0$ . Set it up and solve.

Worked example

Standard case — find p so that $x^2+px+9=0$ has equal roots

“Equal roots” means one repeated root, so need $b^2-4ac=0$ , with $a=1,\ b=p,\ c=9$ .
Write the condition: $p^2-4(1)(9)=0$ .
Tidy and rearrange: $p^2-36=0\Rightarrow p^2=36$ .
Square-root both ways:

Answer

p=6\ \text{or}\ p=-6

Worked example

Exam level — find k so that $y=2x+k$ is a tangent to $y=x^2-4x+1$

Tangent → one repeated root → $b^2-4ac=0$ . Set line = curve: $x^2-4x+1 = 2x+k$ .
Bring everything across to $=0$ : $x^2-6x+(1-k)=0$ , so $a=1,\ b=-6,\ c=1-k$ .
Set discriminant to zero: $(-6)^2-4(1)(1-k)=0$ .
Expand and solve: $36-4+4k=0\Rightarrow 32+4k=0$ .

Answer

k=-8

Show the full working

$x^2-4x+1 = 2x+k$

$x^2-4x-2x+1-k=0$

Collect the $x$ terms:

$x^2-6x+(1-k)=0$

Set the discriminant to zero, minding the signs (since $-4\times(-k)=+4k$ ):

$(-6)^2-4(1)(1-k)=0$

$36-4(1-k)=0\quad\Rightarrow\quad 36-4+4k=0$

$32+4k=0$

Notice we never found where the line touches. The discriminant answered the question on its own. This is the exact step in 9709/13 Jun 2021 Q3.

Where the marks go

Common mistake

A tangent is one repeated root, so

b^2-4ac=0

, not

>0

. And “no real roots” needs a strict

b^2-4ac<0

, not

\le

. The equals case is a repeated root, and that does count as meeting.

Common mistake

Always rearrange to

=0

before reading off

a,b,c

, and mind the sign of

b

when you square it: with

b=-6

b^2=(-6)^2=36

, not

-36

Now you try

Find the set of values of $m$ for which the line $y=mx-3$ does not meet the curve $y=2x^2+5$ . (9709/11 Nov 2020 Q1)

Your turn— tap to reveal the worked answer (9709/11 Nov 2020 Q1)

“Does not meet” → no real roots → $b^2-4ac<0$ .
Set line = curve and rearrange to $=0$ : $2x^2-mx+8=0$ , so $a=2,\ b=-m,\ c=8$ .
Apply the condition: $(-m)^2-4(2)(8)<0 \Rightarrow m^2-64<0$ .
Quadratic inequality; roots $\pm 8$ , negative between them.

Answer

-8<m<8