+964 Find all real values of $p$ such that
(x+1)(x-2p)
has a minimum value of 0 over all real values of $x$.
+135 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)
+129885 (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
