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(B) Given roots are $z_1 = \sqrt{3} + \sqrt{27} = \sqrt{3} + 3\sqrt{3} = 4\sqrt{3}$ and $z_2 = \sqrt{2} + 5i$. Since the coefficients are rational,irr

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$6$

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(B) Given roots are $z_1 = \sqrt{3} + \sqrt{27} = \sqrt{3} + 3\sqrt{3} = 4\sqrt{3}$ and $z_2 = \sqrt{2} + 5i$. Since the coefficients are rational,irrational roots must occur in conjugate pairs. For $z_1 = 4\sqrt{3}$,the conjugate is $-4\sqrt{3}$. For $z_2 = \sqrt{2} + 5i$,the conjugate is $\sqrt{2} - 5i$. However,for a polynomial with rational coefficients,if $\sqrt{a} + \sqrt{b}$ is a root,then $\pm\sqrt{a} \pm \sqrt{b}$ are also roots. Here,$z_1 = 4\sqrt{3}$. The minimal polynomial for $4\sqrt{3}$ over $\mathbb{Q}$ is $x^2 - 48 = 0$,which has roots $4\sqrt{3}$ and $-4\sqrt{3}$. For $z_2 = \sqrt{2} + 5i$,the minimal polynomial over $\mathbb{Q}$ is found by $(x - \sqrt{2})^2 = (5i)^2$,which is $x^2 - 2\sqrt{2}x + 2 = -25$,or $x^2 + 27 = 2\sqrt{2}x$. Squaring both sides gives $(x^2 + 27)^2 = 8x^2$,which is $x^4 + 54x^2 + 729 = 8x^2$,or $x^4 + 46x^2 + 729 = 0$. This polynomial has roots $\pm\sqrt{2} \pm 5i$. Thus,the total degree is $2 + 4 = 6$.

The minimum degree of a polynomial equation with rational coefficients having $\sqrt{3}+\sqrt{27}$ and $\sqrt{2}+5i$ as two of its roots is

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