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avatar+713 

Find all real values of $p$ such that
(x+1)(x-2p)
has a minimum value of 0 over all real values of $x$.

 Jun 9, 2024
 #1
avatar+88 
+1

If an equation has a minimum value of zero, then it only has one solution, meaning that it can be written as (x + y)^2. The only way to write the given equation as (x+y)^2 is (x + 1)(x - 2(-0.5)) = (x + 1)(x - (-1)) = (x + 1)(x + 1) = (x + 1)^2. Therefore, the only real value for p is -(1/2)

 Jun 9, 2024
 #2
avatar+129733 
+1

(x + 1) ( x -2p)

 

x^2 + ( 1 - 2p)x - 2p

 

The x cootdinate of the  vertex is  - [ 1 -2p ] / 2  = [ 2p - 1]  / 2

 

We   want   this point on the graph  ( [ 2p -1 ] /2 , 0 )

 

So

 

( [2p-1]/2 )^2 + (1 - 2p) ([2p-1] /2) -2p =  0

 

p^2 - p + 1/4  + p - 2p^2 -1/2 + p - 2p =  0

 

-p^2 - p -1/4 =  0

 

p^2 + p + 1/4  = 0

 

(p + 1/2)^2  = 0

 

p + 1/2   = 0

 

p  = -1/2

 

cool cool cool

 Jun 12, 2024

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