The biquadratic equation, two of whose roots are $1+\mathrm{i}, 1- \sqrt{2}$ is
- $x^{4} -4\mathrm{x}^{3}+5x^{2}-2x-2=0$
- $x^{4}-4x^{3}-5x^{2}+2x+2=0$
- $x^{4}+4x^{3}-5x^{2}+2x-2=0$
- $x^{4}+4x^{3}+5x^{2}-2x+2=0$
Roots are 1+i, 1-i (conjugate pair) and 1+sqrt(2), 1-sqrt(2). Sum of roots = 4, product of roots = (1+1)(1-2) = -2. The quadratic factors are (x^2 - 2x + 2) and (x^2 - 2x - 1). Multiplying these: x^4 - 2x^3 - x^2 - 2x^3 + 4x^2 + 2x + 2x^2 - 4x - 2 = x^4 - 4x^3 + 5x^2 - 2x - 2 = 0.
Since the coefficients of the biquadratic equation are real complex roots must appear in conjugate pairs and irrational roots involving square roots must include their conjugate surds. The four roots are 1 + i, 1 - i, 1 - sqrt(2) and 1 + sqrt(2). The sum of the roots is (1 + i) + (1 - i) + (1 - sqrt(2)) + (1 + sqrt(2)) = 4 so the x^3 coefficient is -4. The polynomial is (x^2 - 2x + 2)(x^2 - 2x - 1) which expands to x^4 - 4x^3 + 5x^2 - 2x - 2. This results in x^4 - 4x^3 + 5x^2 - 2x - 2 = 0.