how do i solve the following quadratic equation: 9x² - 4 = 0?

9x² - 4 = 0
(3x - 2)(3x + 2) = 0
x = 2/3 or -2/3

x ² = 4 / 9
x = ± (2/3)

9x^2 - 4 = 0
9x^2 = 4
x^2 = 4/9
x = ±√(4/9)
x = ±2/3

9x^2-4, hey! That's a product of sum and difference of two terms! To factor that, get the square root of 9x^2 and 4. The square of 9x^2 is 3x and the square root of 4 is 2. Next, write in binomial form: "the sum of 3x and 2 multiplied by their difference." The factors are (3x + 2)(3x - 2). And so, 3x + 2 = 0, and 3x - 2 = 0.

The values of x are 2/3 and -2/3.

9x^2 -4 = 0

Factor the equation first
(3x-2) (3x+2) = 0

set each to zero
3x-2=0
3x+2=0

3x=2
3x=-2

divide both sides by 3

x=2/3
x=-2/3

insert each number 2/3 and -2/3 into the original formula (9x^2-4=0) to get the final answer.

Hence, 9(2/3)^2 - 4 = 0 or 9(-2/3)^2 - 4 = 0

12

I hope this helps.....

9x ^2-4=0
add 4 to both sides
9x^ = 4
take the square root of both sides
3x=2
divide both sides by 3
x=2/3

9x² - 4 = 0
(3x-2)(3x+2) = 0
If that is the case either (3x-2) or (3x+2) is equal to zero.
Let (3x-2) be zero first of all:
3x-2 = 0
3x = 2
x =2/3
Let (3x+2) be zero now
3x+2 = 0
3x = -2
x = -2/3

x is equal to 2/3 or -2/3 or you can say x = + - 2/3

hope that helps

9x² - 4 = 0
9x² = 4
x² = 4/9
x = Square Root (4/9)
x = (+ or -) 2/3

9x² - 4 = 0
9x²=4
x²=4/9

x=±√(4/9)

x=±(2/3)