CAIE A-Level · Mathematics 9709 · Quadratics

The Discriminant of a Quadratic — CAIE A-Level Pure Mathematics 1 (9709)

7 min readSyllabus 1.1FreeBy Uzair Khan

Syllabus objective

Find the discriminant of a quadratic polynomial ax^2 + bx + c and use the discriminant (e.g. to determine the number of real roots of the equation ax^2 + bx + c = 0). Knowledge of the term 'repeated root' is included.

Introduction

The discriminant is a single expression derived from the coefficients of a quadratic polynomial that instantly reveals how many real roots the corresponding equation has. In the 9709 exam, discriminant questions appear across a wide range of contexts: determining whether a curve intersects a line, proving a quadratic is always positive, and finding unknown constants given information about the nature of roots. Mastering the discriminant is therefore essential, not only as a standalone skill but as a tool woven throughout the Pure Mathematics 1 paper.


Core Concept

Recall from Completing the Square that the general quadratic equation ax2+bx+c=0ax^2 + bx + c = 0 (with a0a \neq 0) can be solved by the quadratic formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The expression under the square root, b24acb^2 - 4ac, controls whether real solutions exist and how many there are. This expression is called the discriminant.

  • If b24ac>0b^2 - 4ac > 0: the square root gives two distinct real values, producing two distinct real roots.
  • If b24ac=0b^2 - 4ac = 0: the square root vanishes, and the ±\pm gives only one value, producing a repeated root (also called an equal root).
  • If b24ac<0b^2 - 4ac < 0: the square root of a negative number has no real value, so there are no real roots.

Geometrically, these three cases describe how the parabola y=ax2+bx+cy = ax^2 + bx + c relates to the xx-axis.


Key Formulae & Definitions

The Discriminant:

Δ=b24ac\Delta = b^2 - 4ac

where ax2+bx+cax^2 + bx + c is a quadratic polynomial with a0a \neq 0.

Classification of roots:

ConditionNature of roots
Δ>0\Delta > 0Two distinct real roots
Δ=0\Delta = 0One repeated (equal) root: x=b2ax = -\dfrac{b}{2a}
Δ<0\Delta < 0No real roots

Note: A repeated root is a single value of xx at which the parabola just touches (is tangent to) the xx-axis.


Worked Examples

Example 1 — Determining the nature of roots

Determine the nature of the roots of 3x25x+4=03x^2 - 5x + 4 = 0.

Step 1: Identify the coefficients.

a=3,b=5,c=4a = 3, \quad b = -5, \quad c = 4

Step 2: Calculate the discriminant.

Δ=b24ac=(5)24(3)(4)=2548=23\Delta = b^2 - 4ac = (-5)^2 - 4(3)(4) = 25 - 48 = -23

Step 3: Interpret the result.

Since Δ=23<0\Delta = -23 < 0, the equation has no real roots. The parabola lies entirely above the xx-axis (as a=3>0a = 3 > 0).


Example 2 — Finding an unknown constant given a repeated root

The equation x2+kx+9=0x^2 + kx + 9 = 0 has a repeated root. Find the possible values of kk.

Step 1: Identify the coefficients.

a=1,b=k,c=9a = 1, \quad b = k, \quad c = 9

Step 2: Set the discriminant equal to zero (condition for a repeated root).

Δ=k24(1)(9)=0\Delta = k^2 - 4(1)(9) = 0
k236=0k^2 - 36 = 0

Step 3: Solve for kk.

k2=36    k=±6k^2 = 36 \implies k = \pm 6

Step 4: State the repeated roots for each case.

  • When k=6k = 6: repeated root at x=62=3x = -\dfrac{6}{2} = -3.
  • When k=6k = -6: repeated root at x=62=3x = -\dfrac{-6}{2} = 3.

Example 3 — Using the discriminant to find a range of values

Find the values of the constant mm for which mx2+6x+3=0mx^2 + 6x + 3 = 0 has two distinct real roots.

Step 1: Identify the coefficients (noting m0m \neq 0 for a quadratic).

a=m,b=6,c=3a = m, \quad b = 6, \quad c = 3

Step 2: Write the condition for two distinct real roots.

Δ>0    3612m>0\Delta > 0 \implies 36 - 12m > 0

Step 3: Solve the inequality.

36>12m    m<336 > 12m \implies m < 3

Step 4: State the full constraint.

Combined with m0m \neq 0: the equation has two distinct real roots when m<3m < 3 and m0m \neq 0.


Common Mistakes & Examiner Pitfalls

  1. Sign errors with bb: The discriminant uses b2b^2, which is always non-negative regardless of the sign of bb. A frequent error is computing b2b^2 incorrectly when bb is negative, e.g. writing (5)2=25(-5)^2 = -25. Always square the full signed value.

  2. Forgetting a0a \neq 0: When mm (or another letter) is the leading coefficient, students often forget that m=0m = 0 must be excluded separately, since the expression would no longer be quadratic.

  3. Using Δ0\Delta \geq 0 instead of Δ>0\Delta > 0: "Two distinct real roots" requires strictly Δ>0\Delta > 0. The boundary Δ=0\Delta = 0 gives a repeated root — a separate case. Read the question wording with care.

  4. Confusing "no real roots" with "no roots": The discriminant only classifies real roots. In 9709 Pure 1, "no real roots" is the correct conclusion; do not write "no solutions" without this qualifier.

  5. Not simplifying Δ\Delta before interpreting: In harder problems, Δ\Delta involves the unknown — students sometimes forget to rearrange the inequality correctly, particularly flipping the inequality sign when dividing by a negative value.


Practice Questions

Q1. Find the discriminant of 2x27x+52x^2 - 7x + 5 and state the nature of the roots of 2x27x+5=02x^2 - 7x + 5 = 0.

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a=2, b=7, c=5a = 2,\ b = -7,\ c = 5

Δ=(7)24(2)(5)=4940=9\Delta = (-7)^2 - 4(2)(5) = 49 - 40 = 9

Since Δ=9>0\Delta = 9 > 0, the equation has two distinct real roots.

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Q2. The equation 4x2+px+25=04x^2 + px + 25 = 0 has a repeated root. Find the possible values of pp and state the corresponding repeated root in each case.

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For a repeated root, Δ=0\Delta = 0:

p24(4)(25)=0    p2=400    p=±20p^2 - 4(4)(25) = 0 \implies p^2 = 400 \implies p = \pm 20
  • p=20p = 20: repeated root x=208=52x = -\dfrac{20}{8} = -\dfrac{5}{2}
  • p=20p = -20: repeated root x=208=52x = -\dfrac{-20}{8} = \dfrac{5}{2}
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Q3. Show that the equation x2+3x+7=0x^2 + 3x + 7 = 0 has no real roots.

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a=1, b=3, c=7a = 1,\ b = 3,\ c = 7

Δ=324(1)(7)=928=19\Delta = 3^2 - 4(1)(7) = 9 - 28 = -19

Since Δ=19<0\Delta = -19 < 0, the equation has no real roots. \square

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Q4. Find the range of values of kk for which x2+kx+k+3=0x^2 + kx + k + 3 = 0 has no real roots.

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For no real roots, Δ<0\Delta < 0:

k24(1)(k+3)<0k^2 - 4(1)(k + 3) < 0
k24k12<0k^2 - 4k - 12 < 0
(k6)(k+2)<0(k - 6)(k + 2) < 0

The quadratic (k6)(k+2)(k-6)(k+2) is negative between its roots:

2<k<6-2 < k < 6
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Q5. The line y=3x+cy = 3x + c is a tangent to the curve y=x2+5x2y = x^2 + 5x - 2. Find the value of cc.

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At a tangency, the line and curve intersect at exactly one point — a repeated root condition.

Set equal:

x2+5x2=3x+cx^2 + 5x - 2 = 3x + c
x2+2x2c=0x^2 + 2x - 2 - c = 0

For a repeated root, Δ=0\Delta = 0:

44(1)(2c)=04 - 4(1)(-2 - c) = 0
4+8+4c=04 + 8 + 4c = 0
4c=12    c=34c = -12 \implies c = -3
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Connections

Prerequisite: Completing the Square — the quadratic formula (from which the discriminant arises) is derived directly by completing the square on ax2+bx+c=0ax^2 + bx + c = 0. Understanding this derivation gives the discriminant its meaning.

Closely linked within Quadratics:

  • Solving Quadratic Equations — the discriminant predicts what the solving process will yield.
  • Quadratic Inequalities — the sign of the discriminant and the factorisation of Δ\Delta as a quadratic in an unknown constant both require inequality-solving techniques.

Later topics that use the discriminant:

  • Coordinate Geometry: Lines and Curves — tangency conditions between lines and curves reduce to Δ=0\Delta = 0 on the resulting quadratic.
  • Functions — determining the range of a quadratic function often involves setting Δ0\Delta \geq 0 and solving for the output variable.
  • Further Pure topics — in complex numbers (A2), the discriminant underpins why complex roots occur in conjugate pairs for real-coefficient polynomials.

Figures

Three parabolas illustrating the three discriminant cases: two x-intercepts (delta > 0), one tangent point (delta = 0), and no x-intercepts (delta < 0).
Figure 1: The three cases of the discriminant. Blue: Δ > 0, two distinct real roots (y = x² − 3x + 1). Green: Δ = 0, one repeated root (y = x² − 2x + 1). Red: Δ < 0, no real roots (y = x² − x + 1).

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Prerequisites: Completing the Square

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