Question

Find all values of p such that 2(x+4)(x-2p) has a minimum value of -18.

(AoPS, Please show steps)

Answer 1

2(x+4)(x-2p) = 2(x^2 + (4-2p)x - 8p)
Completing the square, we have

2(x + (2-p))^2 - 8p - (2-p)^2)
So, at x = p-2, the value is

2(-8p-(2-p)^2) = -2(p+2)^2
If that is -18,
(p+2)^2 = 9
p+2 = ±3
p = -5 or 1

check:

2(x+4)(x+10) = 2(x+7)^2 - 18
2(x+4)(x-2) = 2(x+1)^2 - 18
Both have vertex at y = -18

Answer 2

Why guys?

Answer 3

Okay, i just saw another post u made "Aops". Yes i know who you guys are. don't think i don't. ur math campus or whatever. look at this link: www.jiskha.com/questions/1577346/in-trapezoid-abcd-overline-ab-is-parallel-to-overline-cd-ab-7-units-and

Really AOPS, really? again? and GET THIS: here's a link of AOPS, the same user here, cheating! the very thing they're making a fuss about!

Link:www.jiskha.com/questions/1798144/point-g-is-the-midpoint-of-the-median-xm-of-xyz-point-h-is-the-midpoint-of-xy-and-point

Really this is just disappointing.

Also now that i think about it the real aops wouldn't d something like that, so my conclusion is u are not aops!