MathematicsFoundation20 min read

Quadratic Equations

Standard form, the three methods of solution, and the discriminant

This topic appears in:

01

Standard form

Definition

Quadratic equation — An equation of one variable in which the highest power of that variable is 2. Its standard form is ax² + bx + c = 0, where a, b and c are real numbers and a ≠ 0.

The condition a ≠ 0 is part of the definition. If a were zero the term would vanish and the equation would be linear, not quadratic.

Before applying any method, the equation must be arranged into standard form: all terms on one side, zero on the other, written in descending powers of x.

ax² + bx + c = 0,a ≠ 0standard form
a
coefficient of x², called the leading coefficienta ≠ 0
b
coefficient of x
c
constant termthe value of y where the curve meets the y-axis
Worked example 1

Write 3x(x − 2) = 5 − x in standard form and state a, b and c.

  1. Expand the left side: 3x² − 6x = 5 − x.Remove brackets before collecting terms.
  2. Move every term to the left: 3x² − 6x − 5 + x = 0.Standard form requires zero on the right.
  3. Collect like terms: 3x² − 5x − 5 = 0.−6x + x = −5x.

a = 3, b = −5, c = −5

02

The graph of a quadratic

Definition

Parabola — The curve produced by the graph of y = ax² + bx + c. It is symmetrical about a vertical line through its turning point, called the axis of symmetry.

Every quadratic graph has exactly one turning point. The sign of a decides its type:

The roots of ax² + bx + c = 0 are the values of x where the parabola meets the x-axis. A quadratic therefore has at most two real roots.

ConditionParabola opensTurning point is a
a > 0upwardminimum
a < 0downwardmaximum

Move a through zero: the parabola flips and flattens. Move c: the curve slides vertically, since c is the y-intercept. Move b: the vertex shifts sideways as well as vertically.

03

Methods of solution

Three standard methods are available. All three give the same roots; they differ in speed and in when they can be applied.

MethodUse it whenLimitation
Factorisationthe factors are readily seenonly works for factorisable expressions
Completing the squarethe vertex or the form (x + p)² = q is neededslower; fractions appear when a ≠ 1
Quadratic formulaany quadratic, including those with irrational rootsnone — but slower than factorising
Worked example 2 — factorisation

Solve x² − 7x + 12 = 0.

  1. Find two numbers whose product is c = 12 and whose sum is b = −7.For a = 1 the factors of the constant that add to b give the split directly.
  2. The numbers are −3 and −4.(−3)(−4) = 12 and (−3) + (−4) = −7.
  3. Factorise: (x − 3)(x − 4) = 0.
  4. Apply the zero-product rule: x − 3 = 0 or x − 4 = 0.If a product is zero then at least one factor is zero.

x = 3 or x = 4

04

The quadratic formula

The formula is completing the square carried out once in general, so that the working never has to be repeated. It applies to every quadratic in standard form, whether or not the expression factorises.

x = [ −b ± √(b² − 4ac) ] / 2athe quadratic formula — valid for every quadratic in standard form
±
gives the two roots: one with +, one with −
b² − 4ac
the discriminant, written Δ or Dsee the next section
Worked example 3 — quadratic formula

Solve 2x² + 3x − 2 = 0 using the formula.

  1. Identify the coefficients: a = 2, b = 3, c = −2.The equation is already in standard form.
  2. Compute the discriminant: Δ = 3² − 4(2)(−2) = 9 + 16 = 25.Take care with the signs: −4 × 2 × (−2) is +16, not −16.
  3. Substitute: x = (−3 ± √25) / (2 × 2) = (−3 ± 5) / 4.
  4. Take each sign in turn: x = 2/4 or x = −8/4.

x = ½ or x = −2

05

The discriminant

Definition

Discriminant — The quantity Δ = b² − 4ac, the expression under the square root in the quadratic formula. Its sign determines the nature of the roots without the roots being calculated.

A question that asks how many roots an equation has, or asks for a value of k giving equal roots, or asks whether a line meets a curve, is a discriminant question. In each case the roots themselves are not required.

Value of ΔNature of rootsThe graph
Δ > 0two distinct real rootscuts the x-axis at two points
Δ = 0one repeated real roottouches the x-axis at one point
Δ < 0no real roots (two complex roots)does not meet the x-axis
Δ is a perfect squaretwo distinct rational rootsthe expression factorises over the integers
Worked example 4

Find the value of k for which x² + kx + 9 = 0 has equal roots.

  1. Equal roots require Δ = 0.This is the condition in row two of the table above.
  2. Here a = 1, b = k, c = 9, so Δ = k² − 4(1)(9) = k² − 36.
  3. Set k² − 36 = 0, giving k² = 36.
  4. Take the square root of both sides, keeping both signs.A squared unknown always yields two values; omitting the negative one loses a valid answer.

k = 6 or k = −6

06

Sum and product of the roots

If α and β are the roots of ax² + bx + c = 0, their sum and product can be written directly from the coefficients. Neither root needs to be found first.

These relations are used to check an answer, and to construct a quadratic when its roots are known.

α + β = −b/aαβ = c/ax² − (α + β)x + αβ = 0the second line builds a quadratic from its roots
α, β
the two roots of the equation
−b/a
sum of the rootsnote the negative sign
c/a
product of the roots
Worked example 5

Form the quadratic equation whose roots are 2 and −5.

  1. Sum of roots: 2 + (−5) = −3.
  2. Product of roots: 2 × (−5) = −10.
  3. Substitute into x² − (sum)x + (product) = 0: x² − (−3)x + (−10) = 0.
  4. Simplify the signs.−(−3) = +3.

x² + 3x − 10 = 0

Common mistake: dividing both sides by x

Given x² = 5x, dividing by x gives x = 5 and loses the root x = 0. Division by a variable is only valid when that variable is known to be non-zero. Instead, bring all terms to one side and factorise: x² − 5x = 0 gives x(x − 5) = 0, so x = 0 or x = 5.

Key points

  1. Standard form is ax² + bx + c = 0 with a ≠ 0. Rearrange before doing anything else.
  2. The sign of a decides whether the parabola opens upward (minimum) or downward (maximum).
  3. Three methods of solution: factorisation, completing the square, and the quadratic formula. The formula always works.
  4. The discriminant Δ = b² − 4ac gives the nature of the roots without solving: positive → two real roots, zero → one repeated root, negative → no real roots.
  5. Sum of roots = −b/a; product of roots = c/a. Use these to check answers and to construct equations.
  6. Never divide an equation by the variable — factorise instead, or a root is lost.

Practice questions

6 questions · 26 marks · full working on every one

Try each one on paper first, then open the working. The marks are shown where they are actually awarded, because that is where they are actually lost.

Short questions

3 · 7 marks

Two marks each, in the style of the short-question section of the paper. Answer in two or three lines.

SQ1[2 marks]
State what the discriminant of a quadratic tells you, and write it down.
Model answer

The discriminant is b² − 4ac. It gives the number of real roots: two if positive, one repeated if zero, none if negative.

Examiner tip. Write the expression and the three cases. Either half alone is one mark.

SQ2[3 marks]
Find the value of k for which x² + kx + 9 = 0 has exactly one real root.
Model answer

b² − 4ac = 0 → k² − 36 = 0 → k = ±6

Examiner tip. Both signs count. Giving only k = 6 costs the final mark.

SQ3[2 marks]
Write down the coordinates of the turning point of y = (x − 3)² + 5 and state whether it is a maximum or a minimum.
Model answer

(3, 5), a minimum, because the coefficient of the squared term is positive.

Examiner tip. Completed square form hands you the turning point directly: the x sign flips, the constant does not.

Long questions

2 · 13 marks

Theory and numerical together, as they appear in the long-question section.

LQ1[6 marks]
Solve 2x² − 7x + 3 = 0 by two different methods and show that they agree.
Mark scheme
  1. Factorising: (2x − 1)(x − 3) = 0[1]
  2. x = 1/2 or x = 3[1]
  3. Formula: identifies a = 2, b = −7, c = 3[1]
  4. x = [7 ± √(49 − 24)] / 4[1]
  5. = (7 ± 5)/4[1]
  6. x = 3 or x = 1/2, the same pair[1]

x = 1/2 and x = 3

Examiner tip. When a = 2 the factorisation is (2x − 1)(x − 3), not (x − 1)(2x − 3). Multiply your brackets back out before you commit.

LQ2[7 marks]
A ball is thrown upward. Its height in metres after t seconds is h = 20t − 5t².
  1. Write h in completed square form. [3]
  2. Hence state the greatest height reached and the time at which it occurs. [2]
  3. Find the total time the ball is in the air. [2]
Mark scheme
  1. Takes out the factor: h = −5(t² − 4t)[1]
  2. Completes the square inside: t² − 4t = (t − 2)² − 4[1]
  3. h = −5[(t − 2)² − 4] = 20 − 5(t − 2)²[1]
  4. Greatest height = 20 mthe squared term is zero there[1]
  5. At t = 2 s[1]
  6. Sets h = 0: t(20 − 5t) = 0[1]
  7. t = 4 st = 0 is the moment of throwing[1]

(a) 20 − 5(t − 2)² (b) 20 m at t = 2 s (c) 4 s

Examiner tip. Take the −5 out before completing the square, not after. Trying to complete the square on −5t² directly is where this question is lost.

Exam questions

1 · 6 marks

Multi-part questions with a full mark scheme.

Q1[6 marks]
The line y = x + k is a tangent to the curve y = x² + 3x + 4.
  1. Show that x² + 2x + (4 − k) = 0. [2]
  2. Use the discriminant to find k. [3]
  3. Find the coordinates of the point of contact. [1]
Mark scheme
  1. Equates the two expressions: x + k = x² + 3x + 4[1]
  2. Rearranges to x² + 2x + 4 − k = 0[1]
  3. A tangent means one repeated root, so b² − 4ac = 0[1]
  4. 4 − 4(4 − k) = 0[1]
  5. 4 − 16 + 4k = 0 → k = 3[1]
  6. x² + 2x + 1 = 0 → x = −1, so the point is (−1, 2)[1]

k = 3, touching at (−1, 2)

Examiner tip. "Tangent" is exam shorthand for "discriminant equals zero". The moment you read the word, write that line down.