Syllabus
solve quadratic equations, and quadratic inequalities, in one unknown
 By factorising, completing the square and using the formula.
Solve quadratic equations
To solve quadratic equations, which are equations of the form ax^2 + bx + c = 0
, there are three main methods:
 factorising
 completing the square
 using the quadratic formula.
Factorising
This method works when the quadratic expression can be written as a product of two linear factors.
For example, consider the equation x^2 + 5x + 6 = 0
.

This equation can be factored as
(x+2)(x+3) = 0
. 
Therefore, the solutions are
x = 2
andx = 3
.
Completing the square
This method involves transforming the quadratic expression into a perfect square trinomial and then solving for x
.
For example, consider the equation 2x^2  8x + 5 = 0
.

2x^2  8x + 5 = 2(x^2  4x) + 5

2(x^2  4x + 4  4) + 5 = 2(x2)^2  3 = 0
. 
Solving for
x
, we getx = 2 ± \frac{\sqrt{6}}{2}
.
Quadratic formula
This method involves using the formula x = \frac{b ± \sqrt{b^2  4ac}}{2a}
.
For example, consider the equation 3x^2 + 4x  2 = 0
.
 Using the quadratic formula, we get
x = \frac{4 ± \sqrt{4^2  4 \times 3 \times (2)}}{2 \times 3}
, which simplifies tox = \frac{2 ± \sqrt{7}}{3}
.
Solve quadratic inequalities
To solve quadratic inequalities, which are inequalities of the form ax^2 + bx + c > 0
or ax^2 + bx + c < 0
, we can use a combination of algebraic techniques and graphical analysis.
Algebraic method
We can solve quadratic inequalities by factoring, completing the square, or using the quadratic formula to find the solutions to the related quadratic equation.
We then use these solutions to break the number line into intervals and test a point within each interval to determine the sign of the quadratic expression.
For example, consider the inequality x^2  3x + 2 > 0
.

Factoring the quadratic expression as
(x1)(x2) > 0
, we find that the solutions arex < 1
andx > 2
. 
Testing a point in each interval, we see that the inequality is true for
x < 1
orx > 2
.
Graphical method
We can graph the quadratic expression and use its shape to determine the intervals where it is positive or negative.
For example, consider the inequality x^2  3x + 2 > 0
.

a > 0
, so the parabola opens upwards, i.e. Ushape. 
xintercepts at
x = 1
andx = 2
. 
The region above the parabola is where the expression is positive, so the solution to the inequality is
x < 1
orx > 2
.