How do I solve 81b^2 - 1 =0?

81b^2 - 1 = 0

Since 81 = 9^2

(9b)^2 - 1 = 0

(9b + 1)(9b - 1) = 0

Hence,

9b = -1 or 9b = 1

b = -1/9 or b = 1/9

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

81b^2 - 1 = 0
81b^2 + 9b - 9b - 1 = 0
(81b^2 + 9b) - (9b + 1) = 0
9b(9b + 1) - 1(9b + 1) = 0
(9b + 1)(9b - 1) = 0

9b + 1 = 0
9b = -1
b = -1/9

9b - 1 = 0
9b = 1
b = 1/9

∴ b = ±1/9

1. Factor 81b^2 - 1.
2. The above answer is: (9x+1) (9x-1) If you need help factoring, send me a message or pose a question and I will answer it.
3. Set (9x+1) (9x-1) = to 0
4. That means either (9x+1) or (9x-1) = 0.
5. x either equals 1/9 or -1/9 (answer).

You factor out the problem which turns it into (8b-1)(8b+1)=0. This means 8b-1=0 or 8b+1=0. 8b-1=0 becomes 8b=1 which means b=1/8. 8b+1=0 becomes 8b=-1, which leaves b=-1/8. The total answer is b=+-1/8.

b² = 1/81
b = ± (1/9)