The problem with a quadratic equation occurs when it won’t factorise. But in all cases
we can use the following formula for the solution of the quadratic
ax^2 +bx+c= 0x=−b±√
b^2 − 4 ac
2 aWe will derive this formula below – for now we look at its properties and how to apply it.
First notice that the division byais permissible – ifawere zero then we wouldn’t
have a quadratic. Next notice that the nature of the solutions depends on what is under
the square root, i.e. on the object
=b^2 − 4 acwhich is called thediscriminantof the quadratic. There are three cases:
> 0 ,then√
b^2 − 4 acis a real numberand we have two different real roots.
= 0 ,then√
b^2 − 4 ac= 0and we have two equal real roots.
< 0 ,then√
b^2 − 4 acis a complex numberand we have two different complex (conjugate) roots (see Chapter 12).
In this chapter we are concerned only with the first two cases.
Example
In the equation 3x^2 + 5 x− 2 =0wehavea=3,b=5,c=−2 (always remember to
include signs). So the solutions are
x=− 5 ±√
52 − 4 × 3 ×(− 2 )
2 × 3=− 5 ±√
25 + 24
6=− 5 ± 7
6=2
6or−12
6i.e.x=^13 or−2asbefore.
To see where the formula for the solution of a quadraticequationcomes from we
complete the squareof the quadraticexpression. This procedure has many applications
in elementary and further mathematics and gives good practice in algebraic manipulation.
Basically, it is a technique for expressing the quadratic as a sum or difference of two
squares, and it relies on the key result (42
➤
)
(x+a)^2 =x^2 + 2 ax+a^2 (i)