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Find the value of a for which there is exactly one real value of x such that f(x) = a, where
f(x) = x^2 + 4x - 31 + x^2 + 18x - 45.

 Dec 21, 2023
 #1
avatar+126665 
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f(x) = 2x^2 + 22x - 76

 

2x^2  + 22x - 76    =     a

 

2x^2 + 22x  - 76 - a  =  0

 

2x^2 + 22x - (76 + a) = 0

 

Take the discriminant and set to 0

 

(-22)^2  - 4 (2) [ -76 - a ] = 0

 

484 + 608 + 8a  =0

 

1092  = -8a

 

a = 1092/ -8  =   -136.5

 

(The value of x  =  5)

 

(In other words, f(5)  = -136.5  )

 

 

cool cool cool

 Dec 22, 2023

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