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Let \(f(x) = x^2 + ax + b\) and \(g(x) = x^2 + cx + d\) be two distinct polynomials with real coefficients such that the x-coordinate of the vertex of f is a root of g and the x-coordinate of the vertex of g is a root of f and both f and g have the same minimum value. If the graphs of the two polynomials intersect at the point (100,-100) what is the value of a+c?

 Jan 1, 2020
 #1
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The x coordinate of the vertex of  f  =   -a/2

The x coordinate of the  vertex of g   = -c/2

 

So    (-c/2)  is a root of  f

And (-a/2)  is a root of g

 

So

 

(-c/2)^2  - a(c/2)  +  b   = 0

(-a/2)^2  - c(a/2)  +  d  = 0              simplify

 

c^2/4 - ac/2  + b  = 0

a^2/4   - ac/2  +  d  = 0           multiply both equations through by 4

 

c^2  -2ac  + 4b   =  0

a^2  - 2ac  + 4d  = 0            subtract the second equation  from the  first

 

c^2  - a^2  +  4 ( b - d)  =  0

(c + a) (c - a)  =     4 ( d - b)      (1)  

 

And since  f and  g intersect at  (100, -100)   then

 

100^2  + 100a  + b  =  100^2 + 100c + d

100a + b   =  100c + d         rearrange  as

100(a - c)  = (d - b)        multiply both sides  by 4

400 (a - c)  =  4 (d - b)      (2)

 

Sub (2)   into (1)

 

(c +a)(c - a)  =  400(a - c)        divide both sides by  ( c - a)

 

(c + a)  =   400( a - c)  / ( c - a)

 

(a + c)  =  -400 ( c - a)  / (c - a)

 

a + c   =  -400

 

 

cool cool cool

 Jan 1, 2020
edited by CPhill  Jan 1, 2020
edited by CPhill  Jan 1, 2020

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