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Rewrite the expression 6j^2 - 6j - 12  in the form c(j + p)^2 + q, where c, p, and q are constants. What is q/p?

 Oct 25, 2020
 #1
avatar+9142 
+2

6j2 - 6j - 12

                          Factor  6  out of all the terms

=   6(j2 - j - 2)    (Notice here that if we distributed the 6, we would get back the previous expression.)

 

Now we want to force there to be a perfect square trinomial within the parenthesees. To do that, let's add and subtract half the coefficient of the middle term squared. That is, let's add and subtract  (1/2 * -1)2  which is  (-1/2)2  which is  1/4 . This will create a perfect square trinomial without changing the value of the expression.

 

=   6(j2 - j + \(\frac14\) - \(\frac14\) - 2)

                                       Now there is a perfect square trinomial which is highlighted below:

=   6(j2 - j + \(\frac14\) - \(\frac14\) - 2)

                                       Then  j2 - j + 1/4  factors as  (j - 1/2)2

=   6( (j - \(\frac12\))2  - \(\frac14\) - 2)

                                       And  2 = 8/4

=   6( (j - \(\frac12\))2  - \(\frac14\) - \(\frac84\))

                                       And -1/4 - 8/4  =  -9/4

=   6( (j - \(\frac12\))2  - \(\frac94\) )

                                       Distribute  6  to the terms in parenthesees

=   6( j - \(\frac12\) )2 -  6( \(\frac94\) )

 

=   6( j - \(\frac12\) )2\(\frac{27}{2}\)

 

Now it is in the form  c(j + p)2 + q,  where   c = 6,  p =  -\(\frac12\),  and  q = -\(\frac{27}{2}\)

 

So  q / p   =   ( -\(\frac12\) ) / ( -\(\frac{27}{2}\) )   =   ( -\(\frac12\) ) * ( -\(\frac{2}{27}\) )   =   \(\frac{1}{27}\)

 Oct 25, 2020
 #2
avatar+28021 
+1

Like Hecticar started, factor out 6 from the first two terms

6 (j^2 - j)     - 12      'complete the square' for j

6 (j-1/2)2   - 6/4 - 12

6 (j-1/2)2 - 13 1/2

 Oct 25, 2020

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