A quadratic equation is an
equation where the highest power of x is x². 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, Ö36 = + 6 only.
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. This 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).
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
Where the equation is ax² + bx
+ c = 0
Example
Solve 3x² + 5x  8 = 0
x = 5 ± Ö( 5²  4×3×(8))
6
= 5 ± Ö(25 +
96) 6
= 5 ± Ö(121) 6
\ x = 1 or 2.66
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.
Copyright © Matthew Pinkney 2003
