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Difficulty: Very hardQuadratic Equations and Polynomial Factoring
What is the sum of all real solutions to the polynomial equation (x2−3x+1)2−3(x2−3x+1)+1=x?(x^2 - 3x + 1)^2 - 3(x^2 - 3x + 1) + 1 = x ?

Answer: 6

Answer

The sum of all real solutions to the equation is 6.
The equation (x2−3x+1)2−3(x2−3x+1)+1=x(x^2 - 3x + 1)^2 - 3(x^2 - 3x + 1) + 1 = x represents f(f(x))=xf(f(x)) = x for f(x)=x2−3x+1f(x) = x^2 - 3x + 1. Expanding and setting to zero yields x4−6x3+8x2+2x−1=0x^4 - 6x^3 + 8x^2 + 2x - 1 = 0. Since solutions to f(x)=xf(x) = x (i.e., x2−4x+1=0x^2 - 4x + 1 = 0) also satisfy f(f(x))=xf(f(x)) = x, we can factor out x2−4x+1x^2 - 4x + 1 to get (x2−4x+1)(x2−2x−1)=0(x^2 - 4x + 1)(x^2 - 2x - 1) = 0. Both quadratic factors have positive discriminants (1212 and 88), ensuring four distinct real roots. By Vieta's formulas, the sum of roots from the first quadratic is 4 and from the second is 2, giving a total sum of 6.

Step-by-Step Solution

1
Express the equation as a polynomial in standard form
Expanding (x2−3x+1)2−3(x2−3x+1)+1−x=0(x^2 - 3x + 1)^2 - 3(x^2 - 3x + 1) + 1 - x = 0 yields x4−6x3+8x2+2x−1=0x^4 - 6x^3 + 8x^2 + 2x - 1 = 0.
Expanding allows the composite expression to be treated as a degree-4 polynomial.
2
Find a quadratic factor using fixed points of the iteration
Let f(x)=x2−3x+1f(x) = x^2 - 3x + 1. The equation has the form f(f(x))=xf(f(x)) = x. Any solution to f(x)=xf(x) = x must satisfy this equation. Solving x2−3x+1=xx^2 - 3x + 1 = x gives x2−4x+1=0x^2 - 4x + 1 = 0.
If f(x)=xf(x) = x, then f(f(x))=f(x)=xf(f(x)) = f(x) = x, so x2−4x+1x^2 - 4x + 1 must be a factor of the 4th-degree polynomial.
3
Factor the degree-4 polynomial completely
Polynomial division of x4−6x3+8x2+2x−1x^4 - 6x^3 + 8x^2 + 2x - 1 by x2−4x+1x^2 - 4x + 1 yields (x2−4x+1)(x2−2x−1)=0(x^2 - 4x + 1)(x^2 - 2x - 1) = 0.
Factoring reduces the 4th-degree equation into two quadratic equations.
4
Verify real roots and calculate their sum
For x2−4x+1=0x^2 - 4x + 1 = 0, discriminant D1=12>0D_1 = 12 > 0 and sum of roots is 4. For x2−2x−1=0x^2 - 2x - 1 = 0, discriminant D2=8>0D_2 = 8 > 0 and sum of roots is 2. The four roots are all real and distinct, giving a total sum of 4+2=64 + 2 = 6.
Vieta's formulas provide the sum of roots for each quadratic factor without needing to compute individual radical expressions.

Key Concept

Polynomial Factoring and Composite Quadratic Equations
Estimated Time:2m 30s
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Topics

Quadratic Equations and Polynomial Factoring

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