E) Quadratic Formula & Discriminant Objective: To solve quadratic equations using the quadratic formula and calculate the number of solutions using the discriminant.

Quadratic Formula Discriminant and

Used to determine the # of x-intercepts Discriminant: Used to determine the # of x-intercepts b2 − 4ac = + Two real roots b2 − 4ac = 0 One real root b2 − 4ac = − No real roots No Solution

1) y = x2 + 0x − 4 1 -4 Use the Discriminant to find the # of roots then graph the equation to find the x-intercepts 1) y = x2 + 0x − 4 Discriminant b2 – 4ac = ( )2 – 4( )( ) 1 -4 = + Axis of Symmetry x = -b 2a = −( ) 2( ) = 0 1 Vertex y = ( )2 − 4   = -4 x = {-2, 2}  2nd Point y = ( )2 − 4 Choose x = 1 1 = -3

2) y = x2 + 0x + 0 1 Use the Discriminant to find the # of roots then graph the equation to find the x-intercepts 2) y = x2 + 0x + 0 Discriminant b2 – 4ac = ( )2 – 4( )( ) 1 = 0 x = 0 Axis of Symmetry x = -b 2a   = −( ) 2( ) = 0  1 Vertex y = ( )2 = 0 2nd Point y = ( )2 Choose x = 1 = 0

3) y = x2 + 0x + 4 1 4 Use the Discriminant to find the # of roots then graph the equation to find the x-intercepts 3) y = x2 + 0x + 4 Discriminant b2 – 4ac No x-intercepts    = ( )2 – 4( )( ) 1 4 = − Axis of Symmetry x = -b 2a = −( ) 2( ) = 0 1 Vertex y = ( )2 + 4 = 4 2nd Point y = ( )2 + 4 Choose x = 1 1 = 3

Used to find the x-intercepts Play Video Quadratic Formula Used to find the x-intercepts Axis of Symmetry Discriminant x = -b 2a ± Rules 1) Write problem as a quadratic equation (ax2 +bx + c = 0) 2) Determine values for a, b, and c 3) Plug values into the Quadratic Formula 4) Simplify the Discriminant (factor out “pairs” if possible) 5) Separate into 2 problems (+) and (−) and solve

Example 1 y = -x2 + 5x + 2 - ( ) ± √( )2 − 4( )( ) 2( ) 5 5 -1 2 = -1 b c - ( ) ± √( )2 − 4( )( ) 2( ) 5 5 -1 2 = -1 =

Example 2 y = 2x2 + 5x − 3 - ( ) ± √( )2 − 4( )( ) 2( ) 5 5 2 -3 2 a b c - ( ) ± √( )2 − 4( )( ) 2( ) 5 5 2 -3 2

3 Example 3 y = 3x2 − 4x − 1 - ( ) ± √( )2 − 4( )( ) 2( ) -4 -4 3 -1 = b c - ( ) ± √( )2 − 4( )( ) 2( ) -4 -4 3 -1 = 3 = = = = 3

Example 4 x2 + 3x = 7 1 - ( ) ± √( )2 − 4( )( ) 2( ) 3 3 1 -7 = 1 = a b c - ( ) ± √( )2 − 4( )( ) 2( ) 3 3 1 -7 = 1 =

Example 5 y = -x2 + 4x − 7 - ( ) ± √( )2 − 4( )( ) 2( ) 4 4 -1 -7 = -1 b c - ( ) ± √( )2 − 4( )( ) 2( ) 4 4 -1 -7 = -1 = No Solution

Example 6 Solve a = 1 b = -6 a b c c = 9 = = = 2

1) 2) y = 2x2 − x + 1 x = 3, 1 2 No Solution 3) y = x2 − x − 2 4) Classwork 1) 2) y = 2x2 − x + 1 x = 3, 1 2 No Solution 3) y = x2 − x − 2 4) y = x2 − 2x + 1 x = {-1, 2} x = 1 5) y = 3x2 − 2x + 4 6) y = -x2 + 5x + 1 No Solution