Question

given that y=2x²+px+16 and that y<0 only when 2

Answer 1

Answer:

Since the leading coefficient (the coefficient before x2) a=2>0, the graph of the equation is downward and it is negative between zeros (if any).

If x=2 and x=k are the two zeros of the equation 2x2+px+16=0, then

2x2+px+16=2(x-2)(x-k).

2x2+px+16=2x2-2(k+2)x+4k

From here you find that (equating similar coefficients in both sides)

4k=16

p=-2(k+2)

So, k=4 and p=-12