Solve $x^4-3 x^3+4 x^2-3 x+1=0$
Answer (2)
Divide the equation by x 2 and rearrange terms to get ( x 2 + x 2 1 ) − 3 ( x + x 1 ) + 4 = 0 .
Substitute y = x + x 1 , so x 2 + x 2 1 = y 2 − 2 , and the equation becomes y 2 − 3 y + 2 = 0 .
Solve the quadratic equation for y to get y = 1 and y = 2 .
Solve for x using x + x 1 = y for each value of y , which gives the solutions x = 1 , 1 , 2 1 + i 3 , 2 1 − i 3 . The final answer is 1 , 1 , 2 1 + i 3 , 2 1 − i 3
Explanation
Problem Analysis We are given the quartic equation x 4 − 3 x 3 + 4 x 2 − 3 x + 1 = 0 . Our goal is to find all values of x that satisfy this equation. Notice that the coefficients of the polynomial are symmetric (1, -3, 4, -3, 1). This suggests a method of solution.
Divide by x 2 Since x = 0 is not a solution, we can divide the entire equation by x 2 to obtain x 2 − 3 x + 4 − x 3 + x 2 1 = 0.
Rearrange Terms Now, we rearrange the terms to group the x 2 and x 2 1 terms, as well as the x and x 1 terms: ( x 2 + x 2 1 ) − 3 ( x + x 1 ) + 4 = 0.
Make a Substitution Let's make a substitution to simplify the equation. Let y = x + x 1 . Then y 2 = ( x + x 1 ) 2 = x 2 + 2 + x 2 1 , so x 2 + x 2 1 = y 2 − 2 .
Simplify the Equation Substituting this into our equation, we get ( y 2 − 2 ) − 3 y + 4 = 0 , which simplifies to y 2 − 3 y + 2 = 0.
Solve for y This is a quadratic equation in y . We can factor it as ( y − 1 ) ( y − 2 ) = 0. Thus, the solutions for y are y = 1 and y = 2 .
Solve for x Now we need to solve for x using the substitution y = x + x 1 . This means we need to solve x + x 1 = 1 and x + x 1 = 2 .
For x + x 1 = 1 , we multiply by x to get x 2 + 1 = x , or x 2 − x + 1 = 0 . Using the quadratic formula, we have x = 2 ( 1 ) − ( − 1 ) ± ( − 1 ) 2 − 4 ( 1 ) ( 1 ) = 2 1 ± 1 − 4 = 2 1 ± − 3 = 2 1 ± i 3 . So the solutions are x = 2 1 + i 3 and x = 2 1 − i 3 .
For x + x 1 = 2 , we multiply by x to get x 2 + 1 = 2 x , or 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 complex problems. For instance, analyzing the vibrational modes of a symmetric structure can be simplified by recognizing and exploiting the symmetry in the system's equations. Similarly, in electrical circuit analysis, symmetric circuits can be more easily solved using techniques that take advantage of the symmetry. The method of solving symmetric equations by dividing by a power of the variable and making a substitution is a powerful technique that can be applied in various contexts.
To solve the equation x 4 − 3 x 3 + 4 x 2 − 3 x + 1 = 0 , we simplify it using a substitution. We find the solutions: 1 , 1 , 2 1 + i 3 , 2 1 − i 3 .
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