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# Completing the Square

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

9x2 - 6x ??
9y2 + 12y ?? 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   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
#2
+5

Thank you! Dec 28, 2015
#3
+5   Dec 28, 2015
#4
0

Dec 28, 2015
#5
0

Thanks, Melody.......   Dec 28, 2015