In an equation of the form k^2=ax^2+bx+c, with a>0, the least possible value of k occurs at x=-b/(2a). In the equation \(k = (6x + 12)(x - 8)\), what is the least possible value for k?

UniCorns555 Feb 24, 2024

#1**0 **

First, we expand the product in the given equation: k = (6x+12)(x−8) = 6x2−36x+96.

We see that a=6, and b=−36, so the least possible value of k occurs when x = 2a−b = 2⋅6−(−36) = 3.

Setting x = 3 into the given equation, we see that this value does indeed result in the least possible value of k:

k = 6(3)+(−36)+96 = 150−108 = 42.

Therefore, the answer is 42.

Boseo Feb 25, 2024