how to complete the square?

Question

9x^2 + 6x - 8

How do you complete the square?

Answer 1

[9x² + 6x ] -8 becomes
[9x² + 6x + 1 ] - 8 - 1 add 1 insidebrackets and subtract 1 outsidebrackets so the net result is 0 and expression has not increased nor decreased
factor what is in bracket
(3x + 1)² - 9 now has the square completed
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Answer 2

9x^2 + 6x - 8 = 0
9x^2 + 6x = 8
x^2 + 2x/3 = 8/9
Take 1/2 of 2/3 and square it and add it to both sides:
x^2 +2x/3 +1/9 =8/9 +1/9 = 1
(x+1/3)^2 = 1
x+1/3 = +/- 1 <-- Take sqrt both sides
x = -1/3 +/- 1
x = 2/3 and -4/3

Answer 3

9x^2 + 6x - 8

write 9x^2 + 6x as
(3x +1)^2 -1

9x^2 + 6x - 8 = (3x +1)^2 -1 -8 =
(3x +1)^2 -9

use the identity (a^2 -b^2) = (a - b) (a +b)

to get
{3x +1 -3} {3x +1 +3} = (3x -2)(3x +4)

Answer 4

9x^2+6x-8
=(3x)^2+2x3xx1+1-8-1
=(3x+1)^2-9.ANS.

Answer 5

9x^2+6x=8
x^2+6x=(8/9)
x^2+6x+9=(90/9)
(x+3)^2=(90/9)
x+3=sqrt(90/9)
x+3=sqrt(90)/3
x=(sqrt(90)/3)-3