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# Need help with rewriting expressions

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Rewrite the expression $6j^2 - 4j + 12$ in the form $c(j + p)^2 + q$, where $c$, $p$, and $q$ are constants. What is $\frac{q}{p}$?

Guest Jan 22, 2018

### Best Answer

#1
+93631
+2

Rewrite the expression $$6j^2 - 4j + 12$$ in the form $$c(j + p)^2 + q$$, where c, p, and q are constants. What is $$\frac{q}{p}$$

$$​​​​6j^2-4j+12\\ =6(j^2-\frac{2}{3}j)+12\\ =6(j^2-\frac{2}{3}j+\frac{1}{9})+12-\frac{6}{9}\\ =6(j-\frac{1}{3})^2+11\frac{1}{3}\\ =6(j-\frac{1}{3})^2+11\frac{1}{3}\\ =6(j-\frac{1}{3})^2+\frac{34}{3}\\~\\ p=\frac{-1}{3}\qquad q=\frac{34}{3}\\ \frac{q}{p}=\frac{34}{3}\div\frac{-1}{3}\\ \frac{q}{p}=\frac{34}{3}\times\frac{3}{-1}=-34\\$$

Melody  Jan 22, 2018
#1
+93631
+2
Best Answer

Rewrite the expression $$6j^2 - 4j + 12$$ in the form $$c(j + p)^2 + q$$, where c, p, and q are constants. What is $$\frac{q}{p}$$

$$​​​​6j^2-4j+12\\ =6(j^2-\frac{2}{3}j)+12\\ =6(j^2-\frac{2}{3}j+\frac{1}{9})+12-\frac{6}{9}\\ =6(j-\frac{1}{3})^2+11\frac{1}{3}\\ =6(j-\frac{1}{3})^2+11\frac{1}{3}\\ =6(j-\frac{1}{3})^2+\frac{34}{3}\\~\\ p=\frac{-1}{3}\qquad q=\frac{34}{3}\\ \frac{q}{p}=\frac{34}{3}\div\frac{-1}{3}\\ \frac{q}{p}=\frac{34}{3}\times\frac{3}{-1}=-34\\$$

Melody  Jan 22, 2018

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