IMAT Mathematics · Equations and Inequalities

Quadratic Equations, the Discriminant and the Roots

A quadratic equation is solved by factorising or by the formula; the discriminant says how many real roots it has, and their sum and product can be read off the coefficients.

All Equations and Inequalities notes

Why this matters

The 2018 and 2019 papers asked for the sum of the solutions of an equation that becomes a quadratic once its fractions are cleared. The discriminant also decides the quadratic inequalities asked in 2023 and 2026.

Concept 1 of 2: Solving quadratic equations and the discriminant

If a product is zero, one of its factors must be zero, so a factorised quadratic gives its roots at once. When it will not factorise, the formula always works. The part under the square root, the discriminant, decides everything: positive gives two roots, zero gives one, negative gives none.

Definition

For $ax^2 + bx + c = 0$ with $a \ne 0$ :

Try factorising first. Incomplete forms are quick: $ax^2 + bx = 0$ gives $x(ax + b) = 0$ ; $ax^2 + c = 0$ gives $x^2 = -\frac{c}{a}$ .
Otherwise use the formula below.
Discriminant $\Delta = b^2 - 4ac$ : $\Delta > 0$ two distinct real roots; $\Delta = 0$ one repeated root $x = -\frac{b}{2a}$ ; $\Delta < 0$ no real roots.

Quadratic formula

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

$a, b, c$ the coefficients of $x^2$ , $x$ and the constant
$\Delta$ the discriminant

Worked example

Solve (a)

2x^2 - 7x + 3 = 0

, (b)

x^2 - 4x - 1 = 0

, (c)

3x^2 = 12x

Practice this conceptself-check · 4 quick reps

Never divide both sides by x

Dividing

3x^2 = 12x

x

gives only

x = 4

and loses

x = 0

. Bring everything to one side and factorise out

x

instead.

Concept 2 of 2: Sum and product of the roots

If the roots are

x_1

and

x_2

, the equation is

a(x - x_1)(x - x_2) = 0

. Expanding and matching coefficients shows that the sum and the product of the roots are fixed by

a

b

and

c

, so you can find them without solving.

Definition

For $ax^2 + bx + c = 0$ with real roots $x_1, x_2$ ( $\Delta \ge 0$ ):

Sum $x_1 + x_2 = -\frac{b}{a}$ , product $x_1 x_2 = \frac{c}{a}$ .
A negative product means the roots have opposite signs.
Useful combinations: $x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2$ and $\frac{1}{x_1} + \frac{1}{x_2} = \frac{x_1 + x_2}{x_1 x_2}$ .
The quadratic with roots $s$ and $t$ : $x^2 - (s + t)x + st = 0$ .

Sum and product of the roots

x_1 + x_2 = -\frac{b}{a} \qquad x_1 x_2 = \frac{c}{a}

Worked example

The roots of

2x^2 - 6x - 5 = 0

are

x_1

and

x_2

. Find

x_1 + x_2

x_1 x_2

x_1^2 + x_2^2

and

\frac{1}{x_1} + \frac{1}{x_2}

Practice this conceptself-check · 4 quick reps

The sum of the roots is minus b over a

Expanding

a(x - x_1)(x - x_2)

gives

-a(x_1 + x_2)

as the

x

coefficient, so the sum is

-\frac{b}{a}

, not

\frac{b}{a}

. And check

\Delta \ge 0

first:

x^2 + 2x + 5 = 0

has no real roots, so it has no real roots to add.

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)

Solving quadratic equations and the discriminant
Quadratic formula
$x = \frac{-b \pm \sqrt{\Delta}}{2a}, \qquad \Delta = b^2 - 4ac$
Sum and product of the roots
Sum and product of the roots
$x_1 + x_2 = -\frac{b}{a} \qquad x_1 x_2 = \frac{c}{a}$

Watch out for (2)

Never divide both sides by x
→ Solving quadratic equations and the discriminant
The sum of the roots is minus b over a
→ Sum and product of the roots

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