solve by completing the square.9x^2 - 6x - 35 = 0?

Answer 1

9x^2 - 6x - 35 = 0
=> 9x^2 - 6x +1 -1- 35 = 0
=> ( 3x-1)^2 - 36 = 0
=> ( 3x-1)^2 = 36
=> 3x-1 = +/- 6
=> 3x-1 = 6 OR => 3x-1 = -6
etc
OR

=> ( 3x-1)^2 - 36 = 0
=> (3x-1-6)(3x-1+6) = 0
=> (3x-7)(3x+5) = 0
=> 3x-7 = 0 OR 3x+5 = 0
etc

Answer 2

Hi,
You could go through the traditional way of dividing by 9, but that would involve some nasty fractions. So, probably the easiest way, to look at this is to set it up as a factorization of two squares:
(3x - __ )(3x - __ )
Now, ask yourself what number you put in the blanks so that the sum of the inside and outside terms =
-6x. That is obviously 1.

Now, back to completing the square:
(9x^2 -6x +1) -1 -35 = 0 (We added 1 so we must subtract it.)
(3x-1)² = 36 (We intentionally made a perfect square.)
3x -1 = +- sqrt (36) (Take the square root of both sides.)
3x = 1+- 6 (Add 1 to each side of the equality.)
I'll leave it to you to do the rest of the arithmetic.

Hope this helps.
FE

Answer 3

9x^2 - 6x -35 = 0
adding 35 on both sides
9x^2 - 6x =35
adding 1 on both sides
9x^2 - 6x +1 = 36
(3x - 1)^2 = 36
(3x - 1)^2 = 6^2
3x -1 = +6 or -6
3x = +7 or -5
x = 7/3 or -5/3