Quadratic Equations

Solving a quadratic equation:

Solve 6x2 - 11x + 4 = 0 by
  1. factoring

    2. (2x - 1)(3x - 4) = 0
    2x - 1 = 0   or    3x - 4 = 0
         2x = 1                3x = 4
           x = 1/2               x = 4/3
  2. completing the square
    1. Move the constant term to the right side

      2. 6x2 - 11x = -4
    2. Divide both sides by the quadratic coefficient

      3. x2 - 11 x = -4
              6         6
    3. Take half the middle coefficient and square it; then add it to both sides

      4. x2 - 11 x + 121 = -4 + 121                                 1  11   =   11    112  =  121
              6       144      6    144                                 2   6          12   122      144
    4. Factor the left side and simplify the right side

      5. (x  -  11)2  =    25
               12         144
    5. Take the square root of both sides

      6. x  -  11  =   +5   or  -5
             12        12        12
    6. Solve for x

      7. x = 11 + 5    or    11 - 5
            12   12           12  12

         =  16   or   6
             12        12

        =  4/3 or 1/2

  3. using the quadratic formula:  x = -b + \/(b2 - 4ac)   or   x = -b - \/(b2 - 4ac)

    4.                                                            2a                                    2a

    x = 11 + \/(121 - 4(6)(4))          x = 11 - \/(121 - 4(6)(4))
                         12                                           12

    x = 11 + \/(121 -96)          x = 11 - \/(121 - 96)
                    12                                       12

    x = 11 + \/25          x = 11 - \/25
              12                           12
     

    x = 11 + 5          x = 11 - 5
              12                      12

    x = 16/12          x = 6/12
       = 4/3                 = 1/2

The part under the radical is called the discriminant:

Discriminant is b2 - 4ac

If ax2 + bx + c = 0 where a, b, and c are real numbers, then

If b2 - 4ac < 0, the equation has 2 complex roots.
If b2 - 4ac = 0, the equation has 1 double root.
If b2 - 4ac > 0, the equation has 2 real roots.
          If b2 - 4ac is a perfect square, the equation has 2 rational roots.
          If b2 - 4ac is not a perfect square, the equation has 2 irrational roots.


Problems

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