SOLVING QUADRATIC EQUATIONS


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PART 1 : LECTURE

solving by factoring.

example1:

factor and solve: x2 + 5x = 14
(solving meaning= what values of x will cause each factor to be zero )

example2:

3x3 - 7x = 0

Introduction to solving by completing the square

solve x2 = 25 2 ways:
1st way :  x2 - 15 = 0
2nd way:
x2 = 25    What numbers can be squared to be 25.

solve  :  
x2  = 7

solve: 2
x2  = 20

(x - 2)2 = 25

2 ( x+ 3)2 = 14

solving by completing the square

solving (x - 3 )2 = 7 is easy

but what if the form is expanded? like:

x2 -6x + 9 =7   or

 How to go from this form x2 - 6x + 2 =0  to  
(x - 3 )2 = 7 ?

The way is called completing the square.

1st step :  x2 - 6x  __________= -2   move the constant term to the right hand 

2nd step    ( -6/2)2 = (3)2                               Find half of the coefficient of the middle term and square it                                       
3rd step    
 x2 - 6x + (3)2   = -2 + (3)2      Add this term to both side

4th step      (x-3)2 = 9 - 2       the left side term is a binomial squared

5th step     (x-3)2 = 7      solve    

x - 3 = +/-   sqrt(7)
x = 3 +/- sqrt (7)         that is x= 3 + sqrt(7) or x = 3 - sqrt(7)


solve x2 - 10x + 15 = 0
         

solve x2 - 9x -5 =0


solve 3x2 - 12x + 7


solve 4x2 - 3x -2 = 0


solving by using the quadratic formula

The quadratic can be understood if we follow the same steps.

consider : ax2 + bx + c = 0

Find the quadratic formulas.


Use the quadratic formula to solve: x2 + 5x - 14 =0


PART II - QUIZ

from assignments due today

PART III assignments for next time.

to do.. under construction
















 

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