How do I factor, thus solving:9y^2 -1 =0?

Answer 1

the general rule is that : a^2 - b^2 = (a+b)(a-b)

So, 9y^2 - 1 = (3y+1)(3y-1) = 0

3y = -1 or 3y = 1

y = +/- 1/3

Answer 2

There is a difference between FACTORING and SOLVING an equation. You can SOLVE the equation, as follows:

9y^2 - 1 = 0
9y^2 = 1
y^2 = 1/9
y = +1/3 or -1/3

FACTORING means determining two or more expressions that when multiplied together result in the term on the left side of the equation. The equation can be FACTORED, as follows:

9y^2 -1 = 0
(3y +1) * (3y - 1) = 0

=>

3y + 1 = 0
or
3y - 1 = 0

3y + 1 = 0
3y = -1
y = -1/3

3y - 1 = 0
3y = 1
y = 1/3

Answer 3

This is "The difference of 2 perfect squares."

9y^2 is 3y squared; 1 is 1x1 or 1 squared

(3y + 1)( 3y - 1)

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Answer 4

Here we are asked to factor out 9y^2-1=0.
The 9 can be written as 3^2, then 9y^2-1= 0 becomes
3^2y^2 -1^2 =0
we know that x^2y^2 = (xy)^2
(3y)^2 -1^2 = 0.
By the formula, we can write it as,
(3y+1)(3y-1) =0
Implies that (3y+1) = 0 or (3y-1) =0
Finally we get y= -1/3 or y = 1/3.

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Answer 5

This is a difference of squares, so it factors as
(3y-1)*(3y+1) = 0 and
y = ± 1/3

Doug

Answer 6

9.y² = 1
y² = 1 / 9
y = ± (1/3)
or
(3y - 1).(3y + 1) = 0
y = 1/3 , y = - 1/3 (as above)

Answer 7

9y^2 = 1
y^2 = 1/9
y= (radical) 1/9