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# Rewrite the expression

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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
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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
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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