For k0,if k sqrt{-1} is a root of the equati | vedclass

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(C) Given that $k \sqrt{-1} = ki$ is a root of the equation $x^4+6 x^3-16 x^2+24 x-80=0$. Since the coefficients are real,the conjugate $-ki$ must als

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$4$

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(C) Given that $k \sqrt{-1} = ki$ is a root of the equation $x^4+6 x^3-16 x^2+24 x-80=0$. Since the coefficients are real,the conjugate $-ki$ must also be a root. Thus,$(x-ki)(x+ki) = x^2+k^2$ is a factor of the polynomial. Dividing $x^4+6 x^3-16 x^2+24 x-80$ by $x^2+k^2$: $x^4+6 x^3-16 x^2+24 x-80 = (x^2+k^2)(x^2+6x-(16+k^2)) + (24-6k^2)x + (k^2(16+k^2)-80)$. For $x^2+k^2$ to be a factor,the remainder must be zero. Setting the coefficient of $x$ to zero: $24-6k^2 = 0 \implies 6k^2 = 24 \implies k^2 = 4$. Checking the constant term: $k^2(16+k^2)-80 = 4(16+4)-80 = 4(20)-80 = 0$. Thus,$k^2 = 4$ is the correct value.

For $k>0$,if $k \sqrt{-1}$ is a root of the equation $x^4+6 x^3-16 x^2+24 x-80=0$,then $k^2=$

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