Solve the polynomial equation in the complex numbers.
[tex]12 x^4-40 x^3+23 x^2-x-1=0[/tex]
Find the complex zeros of [tex]f[/tex].
[tex]x=[/tex] _
(Type an exact answer, using radicals as needed. Express complex numbers in terms of [tex]i[/tex]. Use a comma to separate answers as needed.)
Answer (2)
Find the roots of the quartic polynomial equation 12 x 4 − 40 x 3 + 23 x 2 − x − 1 = 0 .
The roots are found to be approximately -0.1667, 0.3820, 0.5, and 2.6180.
Express the roots exactly as x = − 6 1 , 2 1 , 2 3 − 5 , 2 3 + 5 .
The complex zeros of f are x = − 6 1 , 2 1 , 2 3 − 5 , 2 3 + 5 .
Explanation
Problem Analysis We are given the quartic polynomial equation 12 x 4 − 40 x 3 + 23 x 2 − x − 1 = 0 and asked to find its complex roots. Since the coefficients of the polynomial are real, any complex roots must occur in conjugate pairs. We can use a calculator or computer algebra system to find the roots of the polynomial.
Finding the Roots Using a computational tool, we find the roots to be approximately -0.1667, 0.3820, 0.5, and 2.6180. We can express these roots exactly as − 6 1 , 2 3 − 5 , 2 1 , and 2 3 + 5 .
Final Answer Therefore, the roots of the given polynomial equation are x = − 6 1 , 2 1 , 2 3 − 5 , 2 3 + 5 .
Examples
Quartic equations, like the one we solved, appear in various engineering and physics problems, such as determining the stability of structures or analyzing the behavior of complex systems. For instance, when designing a bridge, engineers might use quartic equations to model the stresses and strains on different parts of the structure. By finding the roots of these equations, they can identify potential weak points and ensure the bridge's safety and stability. Similarly, in physics, quartic equations can arise in the study of optics and quantum mechanics, where they help describe the behavior of light and particles.
The complex zeros of the polynomial equation 12 x 4 − 40 x 3 + 23 x 2 − x − 1 = 0 are x = − 6 1 , 2 1 , 2 3 − 5 , 2 3 + 5 . After applying numerical methods or graphing approaches, we find these exact values that serve as the solutions. All roots are real in this case, as no complex conjugates arise from the polynomial evaluation.
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