Matematika Ria
Tambahkan ke favorit Tambahkan ke favorit
link Tautkan di sini
Beranda Matematika RiaBeranda








Hak cipta © 2009
MatematikaRia.com

Quadratic Equation

This is a Quadratic Equation:
Quadratic Equation
(a, b, and c can have any value, except that a can't be 0.)

  • The letters a, b and c are the coefficients (you know these)
  • The letter "x" is the variable or unknown (you don't know it yet)
  • (See Basic Algebra Definitions)

The name Quadratic comes from "quad" meaning square, because the highest exponent is a square (in other words x2).

Examples of Quadratic Equations:

  In this one a=2, b=5 and c=3
     
  This one is a little more tricky:
  • Where is a? In fact a=1, because we don't usually write "1x2"
  • b=-3
  • And where is c? Well, c=0, so is not shown.
  Oops! This one is not a quadratic equation, because it is missing x2 (in other words a=0, and that means it can't be quadratic)

Why is it special?

Quadratic equations can be solved using a special formula called the Quadratic Formula:

Quadratic Formula
The "±" means you need to do a plus AND a minus, and so there are normally TWO solutions !
   

The blue part (b2 - 4ac) is called the discriminant, because it can "discriminate" between the possible types of answer:

  • if it is positive, you will get two solutions
  • if it is zero you get just ONE solution,
  • and if it is negative you get two solutions that include Imaginary Numbers .

Solving

To solve, just plug the values of a, b and c into the Quadratic Formula, and do the calculations.

Example: Solve 5x² + 6x + 1 = 0

Quadratic Formula: x = [ -b ± √(b2-4ac) ] / 2a

Coefficients are: a = 5, b = 6, c = 1

Substitute a,b,c: x = [ -6 ± √(62-4×5×1) ] / 2×5

Solve: x = [ -6 ± √(36-20) ]/10 = [ -6 ± √(16) ]/10 = ( -6 ± 4 )/10

Answer: x = -0.2 and -1

(Check:
5×(-0.2)² + 6×(-0.2) + 1 = 5×(0.04) + 6×(-0.2) + 1 = 0.2 -1.2 + 1 = 0
5×(-1)² + 6×(-1) + 1 = 5×(1) + 6×(-1) + 1 = 5 - 6 + 1 = 0)

Quadratic Equation In Disguise

Some equations may not look like quadratic equations, but with a little clever work they can be made into one:

In disguise What to do In standard form a, b and c
x2 = 3x -1 Move all terms to left hand side x2 - 3x + 1 = 0 a=1, b=-3, c=1
2(x2 - 2x) = 5 Expand (undo the brackets), and move 5 to left 2x2 - 4x - 5 = 0 a=2, b=-4, c=-5
x(x-1) = 3 Expand, and move 3 to left x2 - x - 3 = 0 a=1, b=-1, c=-3
5 + 1/x - 1/x2 = 0 Multiply by x2 5x2 + x - 1 = 0 a=5, b=1, c=-1