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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?

 Feb 24, 2024
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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.

 Feb 25, 2024

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