Question

2х-2х^3 =0 и 2х^3 - 3х^2- 8х+12=0

Answer 1

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Answer 2

To find the solutions to the given equations, we'll set each equation equal to zero and then solve for 'x' in each case.

1. 2x - 2x^3 = 0: Factor out 2x from the left-hand side: 2x(1 - x^2) = 0

Now, we have two factors, and the product of these factors is equal to zero. According to the zero product property, at least one of the factors must be equal to zero. So, we'll set each factor to zero and solve for 'x':

a) 2x = 0 Divide both sides by 2: x = 0

b) 1 - x^2 = 0 Add x^2 to both sides: 1 = x^2

Now, take the square root of both sides: x = ±√1

So, for equation 1, the solutions are x = 0, x = 1, and x = -1.

2. 2x^3 - 3x^2 - 8x + 12 = 0: Unfortunately, this equation cannot be easily factored, so we'll have to use numerical methods to find its solutions. One common numerical method is the Newton-Raphson method.

Let's perform the Newton-Raphson method to approximate the solutions:

Initial guesses: x = -3, x = 0, x = 3

Iteration 1: x = -3 f(x) = 2(-3)^3 - 3(-3)^2 - 8(-3) + 12 = 2(-27) - 3(9) + 24 + 12 = -54 - 27 + 24 + 12 = -45

Iteration 2: x = 0 f(x) = 2(0)^3 - 3(0)^2 - 8(0) + 12 = 0 - 0 - 0 + 12 = 12

Iteration 3: x = 3 f(x) = 2(3)^3 - 3(3)^2 - 8(3) + 12 = 2(27) - 3(9) - 24 + 12 = 54 - 27 - 24 + 12 = 15

The Newton-Raphson method usually converges to the closest root. Based on the above iterations, it appears that the roots are near x = -3, x = 0, and x = 3.

So, for equation 2, the approximate solutions are x ≈ -3, x ≈ 0, and x ≈ 3.

Please note that numerical methods only provide approximations, and these results may not be exact. For more precise solutions, you may need to use advanced numerical techniques or software tools.