Given that y=2x^2 + px +16 and that y<0 only when 2<x<k. Find the value of p and k. [Answer says p is -12 and k is 4]

Question

Pls urgent

Answer 1

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  (which coincides with your answer)