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The parabola with equation y=ax^2+bx+c is graphed below:

The zeros of the quadratic ax^2 + bx + c are at x=m and x=n, where m>n. What is m-n?

 Oct 10, 2019
 #1
avatar+105486 
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We can solve this  for  "a"

 

-3  = a (-4 -2)^2  + 1

 

-3 = a ( -6)^2 + 1

 

-4  = 36a

 

a  = -4/36  = - 1/9

 

So....we have this form

 

y  = (-1/9( ( x - 2)^2  + 1       to find the zeroes....let y  = 0

 

0  = (-1/9)(x - 2)^2 + 1

 

-1  = (-1/9)(x - 2)^2       multiply through by -9

 

9 =  ( x - 2)^2        take both roots

 

±√9  = x - 2

 

±3  =  x - 2    add 2 to both sides

 

±3 + 2  =  x

 

So....the roots are

 

x =  3 + 2  = 5  = m         and   x  =  -3 + 2  =  -1  = n

 

So

 

m - n  =   5 - (-1)   =  6

 

 

 

cool cool cool

 Oct 10, 2019

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