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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

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avatar+91771 
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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 \(\frac{q}{p}\)

 

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

Melody  Jan 22, 2018
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 #1
avatar+91771 
+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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