Solve: 81a² - 4 =0?

it's a perfect square
(9a-2)(9a+2)=0

therefore
9a-2=0
9a=2
a=2/9, a=-2/9

Difference of two squares:

(9a - 2)(9a + 2) = 0

9a ± 2 = 0
9a = ±2
a = ±2/9

81a² - 4 =0 Factor it first:

Take the root of 81 and 4. Place them likewise.
(9a 2)(9a 2)

In a problem such as this in which there are two squared terms separated by a minus sign, the factoring uses one '+' and one '-' to complete the factored equation.

(9a + 2)(9a - 2) Now solve for a in each factored term:

a = -2/9, 2/9

a² = 4/81
a = ± 2/9

see 9a*9a=81a^2
2*2=4
81a² - 4 = 0
=>(9a + 2)(9a - 2) = 0
=>(9a + 2) = 0 or (9a - 2) = 0
a = -2/9 or a = 2/9

a^2 - b^2 = (a + b)(a - b)

81a^4 - 4 = 0
(9a^2 + 2)(9a^2 - 2) = 0

9a^2 + 2 = 0
9a^2 = -2
a^2 = -2/9
a = ±√(-2/9) (imaginary number)

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

∴ a = (±√2)/3

81a² - 4 = 0
81a² = 4
a² = 4/81
a = sq. rt. (4/81)
a = +/-(2/9)

81a^2 = 4

a^2 = 4/81

a = + or - square root 4/81

a = + or - 2/9

81a² - 4 = 0
(9a + 2)(9a - 2) = 0
(9a + 2) = 0 OR (9a - 2) = 0
a = -2/9 OR a = 2/9

+- 2/9