The quadratic \(x^2-6x+66\) can be written in the form \((x+b)^2+c\), where \(b\) and \(c\) are constants. What is \(b+c\)?

Guest Jan 2, 2018

#1**+2 **

One strategy is to rewrite the equation such that to get a perfect square.

\(x^2-6x+\left(\frac{-6}{2}\right)^2\)

What I am doing here is figuring out what number goes after x^2-6x such that it is a perfect square. Simplify from here.

\(x^2-6x+9=(x-3)^2\)

You probably notice that there is a problem, though. The original expression does not end with a term of +9; it ends with a +66. Let's break up the +66.

\(x^2-6x+9+57\)

Notice how breaking up the expression does not actually change its value, so this is a valid step. As aforementioned, the first three terms form a perfect-square binomial.

\((x\textcolor{blue}{-3})^2\textcolor{red}{+57}\\ (x+\textcolor{blue}{b})^2+\textcolor{red}{c}\)

Notice the parallelism in structure. This allows us to identify that -3=b and 57=c. Adding these values together, we get that \(b+c=-3+57=54\)

.TheXSquaredFactor Jan 2, 2018

#1**+2 **

Best Answer

One strategy is to rewrite the equation such that to get a perfect square.

\(x^2-6x+\left(\frac{-6}{2}\right)^2\)

What I am doing here is figuring out what number goes after x^2-6x such that it is a perfect square. Simplify from here.

\(x^2-6x+9=(x-3)^2\)

You probably notice that there is a problem, though. The original expression does not end with a term of +9; it ends with a +66. Let's break up the +66.

\(x^2-6x+9+57\)

Notice how breaking up the expression does not actually change its value, so this is a valid step. As aforementioned, the first three terms form a perfect-square binomial.

\((x\textcolor{blue}{-3})^2\textcolor{red}{+57}\\ (x+\textcolor{blue}{b})^2+\textcolor{red}{c}\)

Notice the parallelism in structure. This allows us to identify that -3=b and 57=c. Adding these values together, we get that \(b+c=-3+57=54\)

TheXSquaredFactor Jan 2, 2018