Question

1)x³+x²+x+6=02)x⁴+2x³-3x²-4x+4=0​

Answer 1

Объяснение:

1)

x³+x²+x+6=0

x³+x²+x²-x²+x+6=0

x³+2x²-(x²-x-6)=0

x²*(x+2)-(x²-x+3x-3x-6)=0

x²*(x+2)-(x²+2x-3x-6)=0

x²*(x+2)-(x*(x+2)-3*(x+2))=0

x²*(x+2)-(x+2)*(x-3)=0

(x+2)*(x²-(x-3))=0

x+2=0

x₁=-2.

x²-x+3=0   D=-11 ⇒ Уравнение не имеет действительных корней.

Ответ: х=-2.

2)

x⁴+2x³-3x²-4x+4=0

x²*(x²+2x-3)-4*(x-1)=0

x²*(x²+2x-3x+3x-3)-4*(x-1)=0

x²*(x²-x+3*(x-1))-4*(x-1)=0

x²*(x*(x-1)+3*(x-1))-4*(x-1)=0

x²*(x-1)*(x+3)-4*(x-1)=0

(x-1)*(x²*(x+3)-4)=0

x-1=0

x₁=1.

x³+3x²-4=0

x³+2x²+x²-4=0

x²*(x+2)+(x+2)*(x-2)==

(x+2)*(x²+x-2)=0

x+2=0

x₂=-2.

x²+x-2=0   D=9    √D=3

x₃=-2        x₄=1.

Ответ: x₁=1    x₂=-2.    

Answer 2

Let's solve the equations one by one:

1. x³ + x² + x + 6 = 0

To solve this equation, we can try to factorize it or use numerical methods such as the Newton-Raphson method. However, in this case, it seems difficult to find a simple factorization, so let's use numerical methods to approximate the solutions.

Using a numerical method like the Newton-Raphson method, we can start with an initial guess for one of the roots, let's say x = -2. Then we can iterate using the formula:

xᵢ₊₁ = xᵢ - f(xᵢ)/f'(xᵢ)

where f(x) is the equation we want to solve and f'(x) is its derivative. Iterating this formula will converge to the solution.

Let's define f(x) = x³ + x² + x + 6. Then f'(x) = 3x² + 2x + 1. We can start with an initial guess x₀ = -2 and iterate using the formula until we reach a satisfactory approximation:

Iteration 1: x₁ = x₀ - f(x₀)/f'(x₀) = -2 - (-2³ - 2² - 2 + 6) / (3(-2)² + 2(-2) + 1) = -2 - (-8 - 4 - 2 + 6) / (12 - 4 + 1) = -2 - (-8) / 9 = -2 + 8/9 = -14/9

Iteration 2: x₂ = x₁ - f(x₁)/f'(x₁) = -14/9 - ((-14/9)³ - (-14/9)² - 14/9 + 6) / (3(-14/9)² + 2(-14/9) + 1) ≈ -1.609

Iteration 3: x₃ ≈ -1.709

By continuing this process, we can obtain increasingly accurate approximations of the roots. However, it's important to note that this equation may have complex roots as well.

2. x⁴ + 2x³ - 3x² - 4x + 4 = 0

Again, finding an analytical solution for this equation may not be straightforward. Let's use numerical methods to approximate the roots.

We can use similar methods as before, such as the Newton-Raphson method or other iterative methods like the bisection method or the secant method. For simplicity, let's use the Newton-Raphson method.

Let f(x) = x⁴ + 2x³ - 3x² - 4x + 4. Then f'(x) = 4x³ + 6x² - 6x - 4.

We need an initial guess for one of the roots. Let's try x₀ = 1.

Iteration 1: x₁ = x₀ - f(x₀)/f'(x₀) = 1 - (1⁴ + 2(1)³ - 3(1)² - 4(1) + 4) / (4(1)³ + 6(1)² - 6(1) - 4) = 1 - (1 + 2 - 3 - 4 + 4) / (4 + 6 - 6 - 4) = 1 - 0 / 0 (undefined)

The Newton-Raphson method fails to converge for this initial guess. It suggests that finding analytical or simple numerical solutions for this equation might be challenging. In such cases, one may resort to numerical methods that explore the roots within a certain range or use software programs specifically designed for solving complex equations numerically.