Question

X^4+4x^3+3x^2+2x-1=0

Answer 1

$x^4+4x^3+3x^2+2x-1=0 \\x^2+3x-1+x^4+4x^3+2x^2-x=0 \\x^2+3x-1+x^4+3x^3-x^2+x^3+3x^2-x=0 \\(x^2+3x-1)+x^2(x^2+3x-1)+x(x^2+3x-1)=0 \\(x^2+3x-1)(x^2+x+1)=0 \\x^2+3x-1=0 \\D=9+4=13 \\x_1= \frac{-3+\sqrt{13}}{2} \\x_2=\frac{-3-\sqrt{13}}{2} \\x^2+x+1=0 \\D=1-4\ \textless \ 0$
Ответ:  $x_1= \frac{-3+\sqrt{13}}{2};\ x_2=\frac{-3-\sqrt{13}}{2}$

Answer 2

To solve the equation x^4 + 4x^3 + 3x^2 + 2x - 1 = 0, we can use various methods. Unfortunately, there is no general algebraic formula to directly solve quartic equations like this one, so we'll need to use numerical methods or factorization techniques. Let's proceed with factoring the equation.

Given: x^4 + 4x^3 + 3x^2 + 2x - 1 = 0

Step 1: Check for possible rational roots using the rational root theorem. The possible rational roots are factors of the constant term (-1) divided by factors of the leading coefficient (1). Possible factors of -1: ±1 Possible factors of 1: ±1

Step 2: Substitute the possible rational roots into the equation and check if any of them satisfy the equation.

If we try x = 1: (1)^4 + 4(1)^3 + 3(1)^2 + 2(1) - 1 = 5 (not equal to 0)

If we try x = -1: (-1)^4 + 4(-1)^3 + 3(-1)^2 + 2(-1) - 1 = -2 (not equal to 0)

No rational roots satisfy the equation, so we need to use other methods to solve it.

One approach is to use numerical methods, such as the Newton-Raphson method or the bisection method, to approximate the solutions. These methods involve iterative calculations and provide numerical approximations rather than exact solutions.

Another option is to use software or calculators with built-in functions for solving equations to find the approximate solutions.

Keep in mind that the given equation may have complex or imaginary solutions as well, as quartic equations can have up to four solutions.