The equation of lowest degree with rational c | vedclass

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(A) Let the roots be $\alpha_1 = \sqrt{3}+\sqrt{2}i$ and $\alpha_2 = \sqrt{3}-\sqrt{2}$.Since the coefficients must be rational,the conjugate of $\alp

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$(x^4-2x^2+25)(x^4-10x^2+1)=0$

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(A) Let the roots be $\alpha_1 = \sqrt{3}+\sqrt{2}i$ and $\alpha_2 = \sqrt{3}-\sqrt{2}$.Since the coefficients must be rational,the conjugate of $\alpha_1$,which is $\bar{\alpha_1} = \sqrt{3}-\sqrt{2}i$,must also be a root.Thus,the quadratic factor corresponding to $\alpha_1$ and $\bar{\alpha_1}$ is $(x-(\sqrt{3}+\sqrt{2}i))(x-(\sqrt{3}-\sqrt{2}i)) = ((x-\sqrt{3})-\sqrt{2}i)((x-\sqrt{3})+\sqrt{2}i) = (x-\sqrt{3})^2 + 2 = x^2-2\sqrt{3}x+5$.Since the coefficients must be rational,we need to eliminate the $\sqrt{3}$ term. The conjugate of $\sqrt{3}-\sqrt{2}$ is $-\sqrt{3}-\sqrt{2}$,and the conjugate of $\sqrt{3}+\sqrt{2}$ is $-\sqrt{3}+\sqrt{2}$.However,for the root $\alpha_2 = \sqrt{3}-\sqrt{2}$,the minimal polynomial over $\mathbb{Q}$ is $(x^2-3)^2 = 2$,which is $x^4-6x^2+9=2$,or $x^4-6x^2+7=0$. Wait,let's re-evaluate.For $\alpha_1 = \sqrt{3}+\sqrt{2}i$,$(x-\sqrt{3})^2 = -2 \implies x^2-2\sqrt{3}x+5=0$. To make coefficients rational,we multiply by the conjugate factor $x^2+2\sqrt{3}x+5=0$. Product: $(x^2+5)^2 - (2\sqrt{3}x)^2 = x^4+10x^2+25-12x^2 = x^4-2x^2+25=0$.For $\alpha_2 = \sqrt{3}-\sqrt{2}$,$(x-\sqrt{3})^2 = 2 \implies x^2-2\sqrt{3}x+1=0$. To make coefficients rational,we multiply by $x^2+2\sqrt{3}x+1=0$. Product: $(x^2+1)^2 - (2\sqrt{3}x)^2 = x^4+2x^2+1-12x^2 = x^4-10x^2+1=0$.The combined equation is $(x^4-2x^2+25)(x^4-10x^2+1)=0$.

The equation of lowest degree with rational coefficients having roots $\sqrt{3}+\sqrt{2} i$ and $\sqrt{3}-\sqrt{2}$ is

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