Question

2x^4-x^3-14x^2+19x-6=0

Answer 1

Ответ: -3; 0,5; 1; 2

Пошаговое объяснение: Решение на фотографии.

Answer 2

The given equation is a polynomial equation. To solve for its roots, we can use various methods, such as factoring, synthetic division, or numerical methods like the Newton-Raphson method. In this case, the polynomial is of degree 4, so finding exact algebraic solutions might be quite complex. We can try to factor it or use numerical methods to approximate the roots.

Let's start by attempting to factor the polynomial:

2x^4 - x^3 - 14x^2 + 19x - 6 = 0

This polynomial doesn't appear to have obvious rational roots, so factoring might not be straightforward. Let's try to use numerical methods to approximate the roots.

We can use a numerical method like the Newton-Raphson method to approximate the roots. The Newton-Raphson method requires an initial guess for each root. Let's start with an initial guess of x = 1 for one of the roots and see if we can find an approximate solution:

1. Choose an initial guess, x_0 = 1.
2. Calculate the next approximation using the formula: x_1 = x_0 - f(x_0) / f'(x_0), where f(x) is the given polynomial and f'(x) is its derivative.

Let's calculate the derivative of the polynomial and then use the Newton-Raphson method to find an approximate root:

Given polynomial: f(x) = 2x^4 - x^3 - 14x^2 + 19x - 6

Derivative: f'(x) = 8x^3 - 3x^2 - 28x + 19

Using the Newton-Raphson formula: x_1 = x_0 - f(x_0) / f'(x_0) x_1 = 1 - (f(1) / f'(1))

Calculating f(1) and f'(1): f(1) = 2(1)^4 - (1)^3 - 14(1)^2 + 19(1) - 6 = 1 - 1 - 14 + 19 - 6 = -1 f'(1) = 8(1)^3 - 3(1)^2 - 28(1) + 19 = 8 - 3 - 28 + 19 = -4

x_1 = 1 - (-1) / (-4) = 1 + 1/4 = 1.25

So, one approximate root is x ≈ 1.25.

You can continue this process with more initial guesses and iterate using the Newton-Raphson method to find other approximate roots. Alternatively, you can use numerical methods or specialized software to find more accurate approximations or even exact solutions if possible.