How do I complete the square and find out the third number in these trinomials?

9x^{2} - 6x ??

9y^{2} + 12y ??

Shades
Dec 28, 2015

#1**+15 **

9x^2 - 6x first, factor out the 9

9 [x^2 - (2/3)x ] Now, inside the brackets, take 1/2 of the coefficient on x = (1/2)(2/3) = 1/3 ....square it .... (1/3)^2 = 1/9.......add this and subtract it.....so we have

9 [ x^2 -(2/3)x + 1/9 - 1/9] factor the 1st three terms

9 [ ( x - 1/3)^2 - 1/9 ] distribute the 9 back across everything

9(x - 1/3)^2 - 1

The second one is similar

9y^2 + 12y

9 [ y ^2 + (4/3)y] ........ (1/2)(4/3) = 2/3 ......(2/3)^2 = 4/9

9[ y^2 + (4/3)y + 4/9 - 4/9] =

9 [ (y + 2/3)^2 - 4/9] =

9(y + 2/3)^2 - 4

CPhill
Dec 28, 2015

#1**+15 **

Best Answer

9x^2 - 6x first, factor out the 9

9 [x^2 - (2/3)x ] Now, inside the brackets, take 1/2 of the coefficient on x = (1/2)(2/3) = 1/3 ....square it .... (1/3)^2 = 1/9.......add this and subtract it.....so we have

9 [ x^2 -(2/3)x + 1/9 - 1/9] factor the 1st three terms

9 [ ( x - 1/3)^2 - 1/9 ] distribute the 9 back across everything

9(x - 1/3)^2 - 1

The second one is similar

9y^2 + 12y

9 [ y ^2 + (4/3)y] ........ (1/2)(4/3) = 2/3 ......(2/3)^2 = 4/9

9[ y^2 + (4/3)y + 4/9 - 4/9] =

9 [ (y + 2/3)^2 - 4/9] =

9(y + 2/3)^2 - 4

CPhill
Dec 28, 2015