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Find all real values of p such that
2(x+8)(x-4p)
has a minimum value of -32 over all real values of x.

 Feb 16, 2024
 #1
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2(x + 8)(x - 4p)

2 (x^2 + (8 -4p)x - 32p)

2x^2 + (16 - 8p)x - 64p      (1)

 

The x coordinate of the vertex is  [ - (16 - 8p) / (2*2) ] =  [8p - 16) /4 = 2p - 4

 

So sub this into this equation

 

2(2p-4)^2 + (16 - 8p)(2p -4) - 64(2p -4)  =  -32

 

Simplifying

 

-8(p^2 + 12p - 28)  = -32

 

p^2 + 12p - 28  = 4

 

p^2 + 12p - 32  =  0

 

p^2 + 12p  =  32

 

p^2 + 12p + 36  = 32 + 36

 

(p + 6)^2  = 68       take both roots

 

p + 6 = sqrt (68)                    p+ 6  = -sqrt (68)

 

p =  2sqrt (17) - 6                  p = -2sqrt(17)  - 6

 

cool cool cool

 Feb 16, 2024

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