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solve ax2+bx+c=0

Guest Feb 21, 2017
 #1
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ax^2+bx+c=0      subtract c from both sides

 

ax^2 + bx  =  -c     divide through by a

 

x^2  + (b/a)x  =   -c/a    take  (1/2) of  (a/b)  =  b / [2a].....square this = b^2/[4a^2]...

add it  to both sides

 

x^2 + (b/a)x + b^2/[4a^2]  = -c/a  + b^2/[4a^2]

 

Factor the left side.....get a common denominator on the right

 

( x  + b/[2a] )^2  =   [ b^2 - 4ac] / [4a^2]

 

Take  both square roots

 

 x  + b/[2a]   =  ±  √  ( [ b^2 - 4ac] / [4a^2]  )

 

 x  + b/[2a]   =  ±  √  ( [ b^2 - 4ac] / [2a]  )     

 

Subtract  b/[2a]  from both sides

 

x  =  ±  √  ( [ b^2 - 4ac] / [2a]  )   -  b/[2a]

 

x =  ( -b  ±  √  ( [ b^2 - 4ac]  )  / [ 2a ]

 

Voila!!!!!......this is  the  Quadratic Formula.....

 

 

cool cool cool

CPhill  Feb 21, 2017

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