find the value of b in the following equation?

Question

Solve for the value of b in the following equation:
3b³ - 5b² = 2b + 0^b

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Answer 1

3b*b*b - 5 b*b = 2b +0^b
3b*b*b - 5b*b - 2b =0
3b*b - 5b - 2 = 0
This is a quadratic equation
now if m = 3, n = 5 c = 2
then b= {-n +-(n^2 - 4mc)^(1/2)}/2m
={-3 +-(9-24)^1/2}/6
={-3+-(15)^1/2}/6
=(0.145497224) or (-1.145497224)

Answer 2

2

Answer 3

The key is to stare at 0^b. When b is negative or zero, this quantity is not defined. (Remember that 1/0 and 0^0 don't make any sense.) Hence we must have b to be positive. In this case, you're asking for solutions to the equation

3 b^3 - 5 b^2 - 2 b = 0

The cubic on the left hand side factors as b (b-2) (3b+1), so that the solutions are b = -1/3, 0, and 2. But remember that b must be positive! So the solution is just b = 2.

Answer 4

after cancelling the common factor b
the quadratic equation is
3(b^2) - 5b -2 = 0
this factorises to
(3b +2 ) ( b-1 ) = 0
therefore the solution is
b = 1, b = -2/3

Answer 5

3b^3 - 5b^2 = 2b + 0
3b^3 - 5b^2 - 2b = 0
b(3b^2 - 5b -2) = 0
b(3b + 1)(b - 2) = 0

therefore,
b = 1
b = -1/3
b = 2

Answer 6

b = 0
b = 1
b = -2/3

Answer 7

2
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Prove (subtitute b with 2):
(3)(2^3) - (5)(2^2) = (2)(2) + (0^2)
(3)(8) - (5)(4) = (4) + (0)
24 - 20 = 4
4 = 4