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# HALP!

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

Jan 2, 2018

### Best Answer

#1
+2341
+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 $$b+c=-3+57=54$$

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Jan 2, 2018

### 1+0 Answers

#1
+2341
+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