Technique

Locate roots by the sign of $f$ at test points

Algebra • Quadratic Equations • Quadratic Inequalities

Statement

Let

f (x) = a x^{2} + b x + c

with

a > 0

and real roots

α \leq β

. Then

f (t) < 0

exactly when

α < t < β

, and

f (t) > 0

exactly when

t < α

t > β

. Hence

p < α < q < β < r

holds precisely when

f (p) > 0

f (q) < 0

and

f (r) > 0

How to use it

1Make the leading coefficient positive first, multiplying the equation by $- 1$ if necessary.
2One root in $(p, q)$ and the other outside it needs only $f (p) f (q) < 0$ .
3Strict inequalities on the roots give strict inequalities on $f$ : a root sitting at an endpoint makes $f$ zero there.
4A sign change already forces two distinct real roots, so no separate discriminant check is needed.

Worked example

For which

m

does

x^{2} - 5 x + m = 0

have one root in

(0, 2)

and the other in

(2, 5)

Step-by-step

1
$f (x) = x^{2} - 5 x + m$ , $a = 1 > 0$
Why: An upward parabola lies below the axis only between its roots.
2
$f (0) > 0 \Rightarrow m > 0$
Why: $0$ is left of both roots.
3
$f (2) < 0 \Rightarrow 4 - 10 + m < 0 \Rightarrow m < 6$
Why: $2$ lies strictly between the roots.
4
$f (5) > 0 \Rightarrow 25 - 25 + m > 0 \Rightarrow m > 0$
Why: $5$ is right of both roots.

Answer

0 < m < 6

Watch out

The inequalities reverse if

a < 0

and this is not normalised first. Forcing BOTH roots into one interval needs more than endpoint signs: also the vertex position and a positive discriminant.

Builds on

Quadratic formula

Also known as

location of roots, sign of a quadratic at a point, interval in which the roots lie

Locate roots by the sign of $f$ at test points

Statement

Worked example

Appears in past papers

2019 (I)

2018 (II)

2018 (I)

2016 (II)