The biquadratic equation,two of whose roots a | vedclass

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(A) Since the coefficients of the biquadratic equation are assumed to be rational,the irrational and complex roots must occur in conjugate pairs. Thus

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$x^4-4 x^3+5 x^2-2 x-2=0$

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(A) Since the coefficients of the biquadratic equation are assumed to be rational,the irrational and complex roots must occur in conjugate pairs. Thus,the roots are $1+i, 1-i, 1-\sqrt{2}, 1+\sqrt{2}$.For the roots $1+i$ and $1-i$:Sum $= (1+i) + (1-i) = 2$Product $= (1+i)(1-i) = 1^2 - i^2 = 1+1 = 2$The quadratic factor is $x^2 - 2x + 2 = 0$.For the roots $1-\sqrt{2}$ and $1+\sqrt{2}$:Sum $= (1-\sqrt{2}) + (1+\sqrt{2}) = 2$Product $= (1-\sqrt{2})(1+\sqrt{2}) = 1^2 - (\sqrt{2})^2 = 1-2 = -1$The quadratic factor is $x^2 - 2x - 1 = 0$.The biquadratic equation is the product of these two factors:$(x^2-2x+2)(x^2-2x-1) = 0$$x^2(x^2-2x-1) - 2x(x^2-2x-1) + 2(x^2-2x-1) = 0$$x^4 - 2x^3 - x^2 - 2x^3 + 4x^2 + 2x + 2x^2 - 4x - 2 = 0$$x^4 - 4x^3 + 5x^2 - 2x - 2 = 0$.

The biquadratic equation,two of whose roots are $1+i$ and $1-\sqrt{2}$,is

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