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?
+9479 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 = ( -\(\frac12\) ) / ( -\(\frac{27}{2}\) ) = ( -\(\frac12\) ) * ( -\(\frac{2}{27}\) ) = \(\frac{1}{27}\)
+37146 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
