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© Matthew Pinkney 2003

Quadratic Equations

A quadratic equation is an equation where the highest power of x is x²., so it is an equation of the form ax² + bx + c = 0. There are various methods of solving quadratic equations, as shown below.

NOTE: If x² = 36, then x = +6 or -6 (since squaring either of these numbers will give 36). However, if we write Ö36 , we usually mean +6 .

Completing the Square

9 and 25 can be written as 3² and 5² whereas 7 and 11 cannot be written as the square of another exact number. 9 and 25 are called perfect squares. Another example is (9/4) = (3/2)². In a similar way, x² + 2x + 1 = (x + 1)².

To make x² + 6x into a perfect square, we add (6²/4) = 9. The resulting expression, x² + 6x + 9 = (x + 3)² and so is a perfect square. The process of making something into a perfect square is known as completing the square. To complete the square in this way, we take the number before the x, square it, and divide it by 4. This technique can be used to solve quadratic equations, as demonstrated in the following example.

Example

Solve x² - 6x + 2 = 0 by completing the square
x² - 6x = -2
[To complete the square on the LHS (left hand side), we must add 6²/4 = 9. We must, of course, do this to the RHS also].
\ x² - 6x + 9 = 7
\ (x - 3)² = 7
[Now take the square root of each side]
\ x - 3 = ±2.646     (the square root of 7 is +2.646 or -2.646)
\ x = 5.646 or 0.354

Completing the square can also be used to find the maximum or minimum point on a graph.

Example

Find the minimum of the graph y = 3x² - 6x - 3 .

In this case, the x² has a '3' in front of it, so we start by taking the three out: y = 3(x² - 2x -1) .  [This is the same, since multiplying it out gives 3x² - 6x - 3]
Now complete the square for the bit in the bracket:
\ y = 3[(x - 1)² - 2]
Multiply out the big bracket:
\ y = 3(x - 1)² - 6

We are trying to find the minimum value that this graph can be. (x - 1)² must be zero or positive, since squaring a number always gives a positive answer. So the minimum value will occur when (x - 1)² = 0, which is when x = 1. When x = 1, y = -6 . So the minimum point is at (1, -6).

The minimum could also have been found by differentiation.

Some people don't like the method of completing the square to solve equations and an alternative is to use the quadratic formula. This is actually derived by completing the square.

The Quadratic Formula

Let’s complete the square in the general case: ax2 + bx + c

Take out a factor of a:
a [ x2 + (b/a)x + (c/a) ]
a [ [x + (b/2a)]2 + (c/a) – (b2/4a2) ]

Hence if ax2 + bx + c = 0,
[x + (b/2a) ]2 = (b2/4a2) – (c/a)
                    =      b2 – 4ac
                                4a2

Now if we take the square root of both sides and simplify, we get the quadratic formula:

Example

Solve 3x² + 5x - 8 = 0

x = -5 ± Ö( 5² - 4×3×(-8))
                   6
  = -5 ± Ö(25 + 96)
               6
  = -5 ± Ö(121)
             6
  = -5 + 11       or        -5 - 11
          6                         6

\ x = 1 or -2.67

Quadratic Functions

Since you only know how to take square roots of positive numbers, the quadratic formula only gives real solutions if b² - 4ac is greater or equal to 0. The expression b² - 4ac is therefore important, and is known as the discriminant

If b2 – 4ac is less than zero, then there are no solutions. This means that there are no values of x giving a value of y of zero, hence the graph of the curve will not cross the x-axis.

If b2 – 4ac = 0 then the quadratic formula says that x = - b/2a, so there is only one solution. The graph will only touch the x-axis at one point, therefore.

However, if b2 – 4ac > 0, there will be 2 solutions to the equation and so the curve will cross the x-axis at 2 points.

Factorising

Sometimes, quadratic equations can be solved by factorising. In this case, factorising is probably the easiest way to solve the equation.

Example

Solve x² + 2x - 8 = 0
\ (x - 2)(x + 4) = 0
\ either x - 2 = 0 or x + 4 = 0
\ x = 2 or x = - 4

If you do not understand the third line, remember that for (x - 2)(x + 4) to equal zero, then one of the two brackets must be zero.