+466 How do I complete the square and find out the third number in these trinomials?
9x2 - 6x ??
9y2 + 12y ??
+128448 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
+128448 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
