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Quadratic Equation and discriminant

This document introduces the quadratic formula as a method for solving quadratic equations. It shows the steps for deriving the formula from completing the square and provides examples of its use. The discriminant is defined as b^2 - 4ac from the quadratic formula. The sign of the discriminant determines the number and type of roots: positive discriminant yields two real roots, zero discriminant yields one real root, and negative discriminant yields two complex roots. Examples are provided to illustrate each case.

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4-8 The Quadratic Formula and the Discriminant p.278-281
We have a number of different way of finding the roots if a quadratic equations #1. Making a table #2. Factoring #3. Completing the Square Now a new way that comes from completing the square. The Quadratic Formula
The Quadratic Formula Solve for x by completing the square. ax 2 + bx + c = 0 ax 2 + bx +  = −c +  b −c 2 x + x += + a a 2 b −c  b   b  x + x+  = +  a a  2a   2a  2 2
The Quadratic Formula Solve for x by completing the square. ax 2 + bx + c = 0 ax 2 + bx +  = −c +  b −c x2 + x +  = + a a 2 b −c  b   b  x2 + x +   = +  a a  2a   2a  b b2 b 2 − 4ac x2 + x + 2 = a 4a 4a 2 2 2 b  b 2 − 4ac  x+  = 2a  4a 2 
The Quadratic Formula Solve for x by completing the square. b b2 b 2 − 4ac x + x+ 2 = a 4a 4a 2 2 2 b  b 2 − 4ac  x+  = 2a  4a 2  b b 2 − 4ac x+ =± 2a 4a 2 −b b 2 − 4ac x= ± 2a 2a − b ± b 2 − 4ac x= 2a
The Quadratic Formula Solve for x by completing the square. − b ± b − 4ac x= 2a 2
How does it work Equation: 3x 2 + 5 x + 1 = 0 a=3 b=5 c =1 − b ± b 2 − 4ac x= 2a
How does it work Equation: 3x + 5 x + 1 = 0 a=3 b=5 c =1 2 − b ± b 2 − 4ac x= 2a x= − ( 5) ± ( 5) 2 − 4( 3)(1) 2( 3) − 5 ± 25 − 12 x= 6 x= − 5 ± 13 − 5 13 = ± 6 6 6
The Discriminant The number in the square root of the quadratic formula. b − 4ac 2 Given x − 5 x + 6 = 0 2 ( − 5) − 4(1)( 6 ) 25 − 24 = 1 2
The Discriminant b − 4ac 2 The Discriminant can be negative, positive or zero If the Discriminant is positive then there are: 2 real answers. If the square root is not a prefect square ( for example 25 ), then there will be 2 irrational roots ( for example 2 ± 5 ).
The Discriminant b 2 − 4ac The Discriminant can be negative, positive or zero If the Discriminant is positive, there are 2 real answers. If the Discriminant is zero, there is 1 real answer. If the Discriminant is negative, there are 2 complex answers. complex answer have i.
Let’s put all of that b information in a chart. Value of Discriminant Type and Number of Roots D > 0, D is a perfect square 2 real, rational roots (ex: x= 2 and x= -4) D > 0, D NOT a perfect square 2 real, Irrational roots (x = √13 x= -√13) D=0 1 real, rational root (double root) (ex: x = 5) D<0 2 complex roots (complex conjugates) (x = 2 ± 3i ) 2 − 4ac Sample Graph of Related Function
Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0
Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0 0, One rational root
Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0 0, One rational root x + 3x + 5 = 0 2 -11, Two complex roots x + 8x − 4 = 0 2 80, Two irrational roots
Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0 0, One rational root x + 3x + 5 = 0 2
Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0 0, One rational root x + 3x + 5 = 0 2 -11, Two complex roots
Describe the roots Tell me the Discriminant and the type of roots x2 + 6x + 9 = 0 0, One rational root x + 3x + 5 = 0 2 -11, Two complex roots x + 8x − 4 = 0 2
Solve using the Quadratic formula x − 8 x = 33 2
Solve using the Quadratic formula x − 8 x = 33 2 x − 8 x − 33 = 0 2 x= − ( − 8) ± ( − 8) − 4(1)( − 33) 2(1) 2
Solve using the Quadratic formula x 2 − 8 x = 33 x 2 − 8 x − 33 = 0 x= − ( − 8) ± ( − 8) 2 − 4(1)( − 33) 2(1) 8 ± 196 8 ± 14 x= = 2 2 8 + 14 22 x= = = 11 2 2 8 − 14 − 6 x= = = −3 2 2
Solve using the Quadratic formula x − 34 x + 289 = 0 2
Solve using the Quadratic formula x − 34 x + 289 = 0 2 x= − ( − 34 ) ± ( − 34) 2(1) 2 − 4(1)( 289 )
Solve using the Quadratic formula x − 34 x + 289 = 0 2 x= − ( − 34) ± ( − 34) 2 − 4(1)( 289) 2(1) 34 ± 1156 − 1156 x= 2 34 ± 0 34 x= = = 17 2 2
Solve using the Quadratic formula x − 6x + 2 = 0 2
Solve using the Quadratic formula x − 6x + 2 = 0 2 x= − ( − 6) ± ( − 6) 2(1) 2 − 4(1)( 2 ) 6 ± 36 − 8 6 ± 28 x= = 2 2 6 2 7 x= ± = 3± 7 2 2
Solve using the Quadratic formula x 2 + 13 = 6 x x 2 − 6 x + 13 = 0 x= − ( − 6) ± ( − 6) 2 − 4(1)(13) 2(1)
Solve using the Quadratic formula x 2 + 13 = 6 x x 2 − 6 x + 13 = 0 x= − ( − 6) ± ( − 6) 2 − 4(1)(13) 2(1) 6 ± 36 − 52 6 ± − 16 x= = 2 2
Solve using the Quadratic formula x 2 + 13 = 6 x x 2 − 6 x + 13 = 0 x= − ( − 6) ± ( − 6) 2 − 4(1)(13) 2(1) 6 ± 36 − 52 6 ± − 16 x= = 2 2 6 ± 4i 6 4 x= = ± i 2 2 2 x = 3 ± 2i