Solve $x^4-3 x^3+4 x^2-3 x+1=0$
Answer (2)
Divide the equation by x 2 and rearrange terms: ( x 2 + x 2 1 ) − 3 ( x + x 1 ) + 4 = 0 .
Substitute y = x + x 1 , leading to y 2 − 3 y + 2 = 0 .
Solve for y , obtaining y = 1 and y = 2 .
Solve for x using x 2 − y x + 1 = 0 , resulting in x = 1 , 1 , 2 1 + i 3 , 2 1 − i 3 .
x = 1 , 1 , 2 1 + i 3 , 2 1 − i 3
Explanation
Problem Analysis We are given the equation x 4 − 3 x 3 + 4 x 2 − 3 x + 1 = 0 . This is a quartic equation, and we need to find its solutions.
Divide by x^2 Notice that the coefficients of the polynomial are symmetric (1, -3, 4, -3, 1). This suggests that we can divide by x 2 and make a substitution to simplify the equation. Since x = 0 is not a solution, we can divide the equation by x 2 .
Result after division Dividing by x 2 , we get x 2 − 3 x + 4 − x 3 + x 2 1 = 0 .
Rearrange terms Rearrange the terms to get ( x 2 + x 2 1 ) − 3 ( x + x 1 ) + 4 = 0 .
Substitution Let y = x + x 1 . Then y 2 = x 2 + 2 + x 2 1 , so x 2 + x 2 1 = y 2 − 2 .
Simplified equation Substitute into the equation to get ( y 2 − 2 ) − 3 y + 4 = 0 , which simplifies to y 2 − 3 y + 2 = 0 .
Solve for y We can factor the quadratic equation as ( y − 1 ) ( y − 2 ) = 0 . Thus, y = 1 or y = 2 .
Solve for x For each value of y , solve the equation x + x 1 = y for x . This is equivalent to solving x 2 − y x + 1 = 0 .
Solve for x when y=1 If y = 1 , then x 2 − x + 1 = 0 . Using the quadratic formula, x = 2 ( 1 ) − ( − 1 ) ± ( − 1 ) 2 − 4 ( 1 ) ( 1 ) = 2 1 ± 1 − 4 = 2 1 ± i 3 . So, x = 2 1 + i 3 or x = 2 1 − i 3 .
Solve for x when y=2 If y = 2 , then x 2 − 2 x + 1 = 0 . This factors as ( x − 1 ) 2 = 0 , so x = 1 is a repeated root.
Final Answer Therefore, the solutions to the equation x 4 − 3 x 3 + 4 x 2 − 3 x + 1 = 0 are x = 1 , 1 , 2 1 + i 3 , 2 1 − i 3 .
Examples
Symmetric equations appear in various fields, such as physics and engineering, where symmetry often simplifies the analysis of complex systems. For instance, in circuit analysis, symmetric components can simplify the calculation of currents and voltages. Similarly, in structural mechanics, symmetric loading conditions on symmetric structures can lead to easier solutions. The technique of exploiting symmetry to simplify equations is a powerful tool in mathematical modeling.
The solutions to the equation x 4 − 3 x 3 + 4 x 2 − 3 x + 1 = 0 are found using substitution and are x = 1 , 1 , 2 1 + i 3 , 2 1 − i 3 . By recognizing the polynomial's symmetric nature, simplifications and substitutions lead to quadratic equations for easy solving. The final solutions include both real and complex roots.
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