Question 1209309
.
Find all real values of p such that  2(x+6)(x-p)
has a minimum value of -4 over all real values of x.
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<pre>
Expression  2(x+6)(x-p)  represents a parabola opened upward
with the vertex  at  x= {{{(-6+p)/2}}},  half-way between x-intercepts -6 and p.


The y-value at the vertex is  

    {{{2*((-6+p)/2+6)*((-6+p)/2-p)}}} = {{{2*((6+p)/2)((-6-p)/2)}}} = {{{-(1/2)*(6+p)^2}}}.


We want

    {{{-(1/2)*(6+p)^2}}} = -4.


It gives  

    {{{(6+p)^2}}} = 8

    6 + p = +/- {{{sqrt(8)}}}

    p = -6 +/- {{{2*sqrt(2)}}}.


<U>ANSWER</U>.  There are two real numbers "p" satisfying the imposed requirements.

         They are  -6 - {{{2*sqrt(2)}}}  and  -6 + {{{2*sqrt(2)}}}.
</pre>

Solved.