2. Getting Started The objective is to be able to solve any quadratic equation by using the quadratic formula. Quadratic Equation - An equation in x that can be written in the standard form ax 2 + bx + c = 0, a  0. Discriminant - The expression b2 b2 - 4ac, for a quadratic equation ax 2 + bx + c = 0.

3. How Many Solutions Will There Be? To determine how many solutions the quadratic equation will have, determine the value of the discriminant.  If b2 b2 - 4ac is positive, then the equation has two solutions.  If b2 b2 - 4ac is zero, then the equation has one solution.  If b2 b2 - 4ac is negative, then the equation has no real solution.

4. How Many Solutions Are There? How many solutions do each of the following equations have? 25  169  )2)(2(434 22  acb Since b 2 - 4ac is positive, there will be two solutions to the equation. 420  16  bac 22 42415  ()()() Since b 2 –4ac is negative, there will be no real solution to the equation.

5. Use the quadratic formula to solve the following equation. a x bbac   2 4 2 x   63628 2 x   68 2 x   2 66417 21 ()()()() () x   68 2 2     x  6 42 2 2 x  622 2  x32 x   68 2     x  6 2 42 2 2 x  622 2 x  32

6. Determine how many solutions each equation has by using the discriminant. Use the quadratic formula to solve each equation.

7. 16  0  bac 22 44422  ()() 1644  () Since the value of the discriminant is zero, there is only one solution to this equation.

8. 416  12  bac 22 42414  ()()()  444  () Since the value of the discriminant is negative, there is no real solution to this equation.

9.  2   bac 22 414 1 3      16  7  123  () Because the value of the discriminant is positive, this equation will have two solutions.

10.   2440 10  244 10 5      1 411 10  1 5 211 10    bbac a 22 4 2 22452 25 ()()()() () 111  y  5 or

11.   440 4 4    4  10  4 42   4 4   41624 222a    bbac 22 444423()() ()  1 10 2 or