Quadratic Equation

A quadratic function is an equation where the exponents of the variables are not greater than 2 but at least one variable with an exponent of 2, and if the variables are multiplied, not greater than 2.

For example: y = x + 2x² + 2

and: xy = 15                       (add the exponents of x and y, 1+1 = 2)

but not: x²y = 15                (variables multiplied so 2 + 1 = 3 > 2, not quadratic)

and not: y = x⁴ + x² + 1       (power of 4 > 2, not quadratic)

We want to concentrate on the following function:

                              y = ax² + bx + c

where s is the unknown and a, b, and c are constants. We'll be interested in solving this equation for those values of x that give a 0 for the value of y, and

in this case we call this equation the quadratic equation:

                      0 = ax² + bx + c

Let's consider what we have for various values of a, b, and c.

If a = 0, then we have y = bx + c, a linear equation. So, we'll be interested only in the cases where a ≠ 0.

If b is zero, then we have y = ax² + c. Now setting y to 0 we have

                                  0 = ax² + c.

Solving for x we have

Since the highest exponent of x is 2, we expect 2 solutions. The two solutions are the positive square root of -c/a and the negative square root of -c/a. It is convenient to show this fact using the symbol ± in front of the square root sign.

You may ask, well, how do we take a square root of a negative number? We'll cover that later. For now, if a and c have opposite signs then we'll have the square root of a positive number.

Example:

Now, suppose c is zero and b is not zero, then we have y = ax² + bx.

Setting y to 0 we have:

                                   0 = ax² + bx

Factoring x we have:

                                  0 = x·•(ax + b)

Now we have two factors whose product is 0. One or the other must be zero, otherwise their product is not zero. Looking at this equation,

        either x = 0 or ax + b = 0.          

Solving for x we have       x = -b/a.

So, the solution is x = 0 or x = -b/a, and we use a solution set for this, like so:

                                    x є {0, -b/a}

Considering the case where a, b, and c are all nonzero,  we can factor the expression or use the quadratic formula.

Example:

With this example, we could factor is and get the same answer. Recall that

               (x + d)·(x + e) = x² + (d+e)·x + d·e

To factor the equation we need two numbers d and e that when multiplied gives us 4 and when added gives us 5, i.e.,  d·e = 4 and (d+e) = 5. So, the factors of 4 are 2·2 and 4·1. Right away we see that 4 + 1 = 5 and we factor

                         x² + 5x +4 into (x+1)(x+4)

             so we have (x+1)(x+4) = 0   which implies

                         (x+1) = 0 or (x+4) = 0

             the solution is          x є {-1, -4}

The Discriminate

In the quadratic formula, the difference: b² – 4ac is called the discriminate of the quadratic equation.

If this discriminate is equal to zero then there is only one solution to the quadratic equation, namely -b/4ac.

If this discriminate is negative then there are no real solutions to the quadratic equation.

Otherwise there are two solutions to the equation.

Derivation of the quadratic equation:

The quadratic formula comes from a technique called “completing the square.”

                          given 0 = ax² + bx + c    (equation 1)

our goal is to rewrite it so that we can factor it into something similar to (x + d)².

          First of all recall that      (x + d)² = x² + 2dx + d²

   so, divide equation 1 by 'a' to make the coefficient of x² equal to 1.

  and we get          (equation 2)

  Compare this equation to

                             x²   +   2dx   +  d²

      notice that

      and,                                

Now, adding this term to both sides of equation 2, we get

                 

     Rearranging the equation we have

Now we have the equation in the right form to factor (the left side) and combine

terms (on the right side) like this:

               or

Now taking the square root of both sides of the equation we get

and  subtracting from both sides we get

finally, combining fractions we have the result:

This is another FREE Algebra PRINTABLE presented to you from the Algebra section of K12math.com