If we express \(3x^2 - 6x - 2\) in the form \(a(x - h)^2 + k\), then what is \(a + h + k\)?

Guest Jan 2, 2018

#1**+2 **

In order to solve this problem, I would use a similar strategy as I used here https://web2.0calc.com/questions/halp_34

There is a slight difference from this problem and the one I hyperlinked, though. The coefficient of the quadratic term is not one. In order to make it one, let's factor it out without changing the value of the expression.

\(3x^2-6x-2\Rightarrow3(x^2-2x)-2\)

Now, the problem is more or less identical to the one I hyperlinked you to. Let's find the value that we can use to manipulate the expression. This magical value is found exactly the same way as completing the square.

\(x^2-2x+\left(\frac{-2}{2}\right)^2\\ x^2-2x+1\\ (x-1)^2\)

This process shows that we need 1 to be inserted inside the parentheses.

\(3(x^2-2x+1)-2-3(1)\)

Since we added one inside the parentheses, we must compensate for that by subtracting the same amount somewhere else in the expression. We need to multiply by three because of the distributive property. Simplify from here and convert into the desired form.

\(3(x^2-2x+1)-2-3(1)\\ 3(x-1)^2-2-3\\ 3(x-1)^2-5\)

Notice the parallelism.

\(\textcolor{green}{3}(x-\textcolor{blue}{1})^2+(\textcolor{red}{-5})\\ \textcolor{green}{a}(x-\textcolor{blue}{h})^2+\textcolor{red}{k}\)

Just like before, a=3, h=1, and k=-5. Therefore, the sum of these values is \(a + h + k=3+1-5=-1\)

TheXSquaredFactor
Jan 2, 2018