CDS Mathematics · Quadratic Equations

Maximum and Minimum of a Quadratic

Completing the square shows the least value of ax² + bx + c when a > 0, or the greatest when a < 0, and where it occurs.

All Quadratic Equations notes CDS Mathematics strategy

Why this matters

Five PYQs, all EASY or MODERATE, and all the same move: complete the square. The vertex is at x = −b/2a, so the question 'for which x is it least' needs no square at all.

Concept 1 of 2: Completing the square

Write the expression as a constant plus or minus a square. A square is never negative, so the constant is the least value (or the greatest, if the square is subtracted), reached where the square is zero.

Definition

$ax^2 + bx + c = a\left(x + \dfrac{b}{2a}\right)^2 + \dfrac{4ac - b^2}{4a}$ .
If $a > 0$ : least value $\dfrac{4ac - b^2}{4a}$ at $x = -\dfrac{b}{2a}$ .
If $a < 0$ : greatest value $\dfrac{4ac - b^2}{4a}$ at the same $x$ .
A word problem ('a number plus four times its square') becomes an expression in one variable first.

Vertex of a quadratic

x = -\dfrac{b}{2a}, \qquad \text{extreme value} = \dfrac{4ac - b^2}{4a}

Worked example

Find the least value of

3x^2 - 6x + 7

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2018 · CDS (II) 2018 — Elementary Mathematics · Q21Easy

Example 1 · Quadratic Equations · Maximum and Minimum of Quadratic Expressions

The minimum value of the expression

2x^2 + 5x + 5

Take out a before completing

2x^2 + 8x + 1

, factor

2

from the

x

terms first:

2(x + 2)^2 - 7

. Completing

x^2 + 8x

inside without the

2

gives the wrong vertex.

Drill 3 more on completing the square

Concept 2 of 2: Extremes of a reciprocal

If the denominator is always positive, the fraction

\dfrac{1}{\text{denominator}}

is largest exactly where the denominator is smallest.

Definition

For $\dfrac{1}{ax^2 + bx + c}$ with $a > 0$ and $b^2 - 4ac < 0$ : the greatest value is $\dfrac{1}{\text{least value of the denominator}} = \dfrac{4a}{4ac - b^2}$ .
It has no least positive value: as $x$ grows the fraction tends to $0$ .
If the denominator can be zero, the fraction has no greatest value.

Greatest value of a reciprocal

\max \dfrac{1}{ax^2 + bx + c} = \dfrac{4a}{4ac - b^2} \quad (a > 0,\ b^2 < 4ac)

Worked example

Find the greatest value of

\dfrac{1}{x^2 - 2x + 5}

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2019 · CDS (I) 2019 — Elementary Mathematics · Q93Moderate

Example 2 · Quadratic Equations · Maximum and Minimum of Quadratic Expressions

What is the maximum value of the expression

\frac{1}{x^2 + 5x + 10}

Invert the value, not the expression

The least denominator is

\dfrac74

for

x^2 + 3x + 4

, so the greatest fraction is

\dfrac47

. An option

\dfrac74

is the denominator's value, not the answer.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Completing the square
Vertex of a quadratic
$x = -\dfrac{b}{2a}, \qquad \text{extreme value} = \dfrac{4ac - b^2}{4a}$
Extremes of a reciprocal
Greatest value of a reciprocal
$\max \dfrac{1}{ax^2 + bx + c} = \dfrac{4a}{4ac - b^2} \quad (a > 0,\ b^2 < 4ac)$

Watch out for (2)

Take out a before completing
→ Completing the square
Invert the value, not the expression
→ Extremes of a reciprocal

Test yourself on Quadratic Equations

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Quadratic Equations chapter test Drill all 5 questions