SOLUTION: Find all values of $p$ such that$$2(x+4)(x-2p)$$has a minimum value of $-18$ over all real values of $x$. (In other words, we cannot have $x$ be nonreal.)
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Quadratic Equations and Parabolas
-> SOLUTION: Find all values of $p$ such that$$2(x+4)(x-2p)$$has a minimum value of $-18$ over all real values of $x$. (In other words, we cannot have $x$ be nonreal.)
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Question 1040895: Find all values of $p$ such that$$2(x+4)(x-2p)$$has a minimum value of $-18$ over all real values of $x$. (In other words, we cannot have $x$ be nonreal.) Answer by robertb(5830) (Show Source):
You can put this solution on YOUR website! The x-value that will give the minimum value will just be the average of the two roots, namely -4 and 2p. Their average is .
===> ,
after direct substitution into the equation.
<===> (p+2)(-p-2) = -9 <===> ===> p+2 = 3, or p+2 = -3
==> p = 1, or p = -5.
