Quadratic equations are equations in which the highest power of the variable is 2, e.g. x2 + 3x + 5 = 0.The difference between an equation and an expression is that an equation has an ‘equal sign’ whilst an expression has no ‘equal sign’.

• x2 + 3x + 5 = 0 is an equation
• x2 + 3x + 5 is an expression.
• (x + 1)(x + 2) = 0 is a quadratic equation because it reduces to x2 + 3x + 2 = 0.

Quadratic equations can be solved in a number of ways, for example:

• by factorisation
• by completing the square
• by using the quadratic formula
• by graphical method
• e.t.c

#### Solving quadratic equations by factorisation

For this method, I assume you are good at factorisation of quadratic expressions. Lets first look at a few examples that involve quadratic equations that are already factorised.

Solve the following equations:

Example 1

(a – 3)(a + 5) = 0
(a – 3)(a + 5) means we are multiplying the two brackets. If the product of two terms is 0, it means either of the two terms is 0.
Therefore either (a – 3) = 0 or (a + 5) = 0.
Which means a – 3 = 0
a = 3
or
a + 5 = 0
a = -5

a = 3 or -5

Example 2

(b – 2)(b – 1) = 0
Either b – 2 = 0
b = 2
or
b – 1 = 0
b = 1

b = 2 or 1

Example 3

y(y – 5) = 0
Either y = 0
or y – 5 = 0
y = 5

y = 0 or 5

Here is something you might have noticed from the three examples above.

• the equation has to be equal to 0
• the equation produces 2 values for the answer. (The values of an equation are called roots).

In all the examples above, the roots for each equation were distinct (different).

Example 4

(x – 5)2 = 0
it is the same as
(x – 5)(x – 5) = 0
Either x – 5 = 0
x = 5
or x – 5 = 0
x = 5

x = 5 or 5
or you can write it as x = 5 twice.

Example 4 is an example of a quadratic equation whose roots are equal.

Example 5

(2x – 8)(x – 5) = 0
Either 2x – 8 = 0
2x = 8
x = 4
or x – 5 = 0
x = 5

x = 4 or 5

If the equation is not factorised you would have to factorise it on your own.

Example 6

e2 + 2e – 8 = 0
Factorise the left hand side
e2 -2e + 4e – 8 = 0
e(e – 2) + 4(e – 2) = 0
(e + 4)(e – 2) = 0
Either e + 4 = 0
e = -4
or e – 2 = 0
e = 2

e = -4 or 2

Example 7

w2 + 5w + 6 = 0
w2 + 2w + 3w + 6 = 0
w(w + 2) + 3(w + 2) = 0
(w + 2)(w + 3) = 0
either w + 2 = 0
w = -2
or
w + 3 = 0
w = -3

w = -2 or -3

Example 8

x2 – 4 = 0
(x + 2)(x – 2) = 0
either x + 2 = 0
x = -2
or x – 2 = 0
x = 2

x = -2 or 2

Example 9

r2 – 25 = 0
(r + 5)(r – 5) = 0
either r + 5 = 0
r = -5
or
r – 5 = 0
r = 5

r = -5 or 5

Example 9

x2 – 4x = 0
x(x – 4) = 0
either x = 0
or x – 4 = 0
x = 4

x = 0 or 4

As you noticed, in all our examples above, the right hand side value was 0. This is not always the case. However if the right hand side is not equal to 0, you first have to make it 0 before you factorise.

Example 10

a2 + a = 90
first transfer 90 from the right hand side to the left hand side
a2 + a – 90
then solve as we did in the examples above.

#### Solving quadratic equations by completing the square

When solving quadratic equations by completing the square we first have to ensure that the left hand side of the equation is a perfect square. Before we go any further, lets look at what a perfect square is like. After completing the square, the quadratic equation is left in this state, (x – 2)2 = 9.

Lets look at a couple of examples that are already perfect squares.

Example 1

(x – 2)2 = 9
since the left hand side is a perfect square, we find the square root of both sides.
(x – 2) = +/-3
the square root of 9 = 3 or -3, hence the +/- there. (3)2 = 9 and (-3)2 = 9.
either x – 2 = 3
x = 2 + 3
x = 5
or
x – 2 = -3
x = 2 – 3
x = -1

x = 5 or -1

Now lets look at the anatomy of a perfect square. This will make it easier for us to understand completing the square.

(x + a)2 = 0
(x + a)(x + a) = 0
x2 + ax + ax + a2 = 0
x2 + 2ax + a2 = 0
lets now assume that x is a variable and a is a constant.
x2 + 2ax + a2 = 0
this now means that 2a is a coefficient of x and a2 is a standalone constant.
the constant a2 is the square of half the coefficient of x. What this means is if the left hand side is to be made a perfect square, you take the coefficient of x, which is 2a, divide it by 2 to get a, then square it to get a2, which is your constant.

Now lets solve the following by completing the square:

w2 + 5w + 6 = 0
first lets transfer 6 to the right hand side of the equal sign, in order to create space for the constant that will make the left hand side a perfect square.
w2 + 5w = -6
now lets add the square of half the coefficient of x to both sides
w2 + 5w + (2.5)2 = -6 + (2.5)2
factorise the left hand side
(w + 2.5)2 = -6 + 6.25
(w + 2.5)2 = 0.25
now the left hand side is a perfect square. Find the square root of both sides.
w + 2.5 = +/-0.5
w = -2.5 +/-0.5
either w = -2.5 + 0.5
= -2
or w = -2.5 – 0.5
= -3

w = -2 or -3