If the bi-quadratic equation f(x)=x^4+2x^3-16 | vedclass

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(A) Given the equation $f(x)=x^4+2x^3-16x^2-22x+7=0$. Since the coefficients are rational,if $2+\sqrt{3}$ is a root,its conjugate $2-\sqrt{3}$ must al

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$3-\sqrt{2}$

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(A) Given the equation $f(x)=x^4+2x^3-16x^2-22x+7=0$. Since the coefficients are rational,if $2+\sqrt{3}$ is a root,its conjugate $2-\sqrt{3}$ must also be a root. Let the four roots be $2+\sqrt{3}, 2-\sqrt{3}, \alpha, \beta$. From the sum of roots: $(2+\sqrt{3})+(2-\sqrt{3})+\alpha+\beta = -2 \implies 4+\alpha+\beta = -2 \implies \alpha+\beta = -6$. From the product of roots: $(2+\sqrt{3})(2-\sqrt{3}) \alpha \beta = 7 \implies (4-3) \alpha \beta = 7 \implies \alpha \beta = 7$. Forming a quadratic equation for $\alpha$ and $\beta$: $t^2 - (\alpha+\beta)t + \alpha \beta = 0 \implies t^2+6t+7=0$. Solving for $t$: $t = \frac{-6 \pm \sqrt{36-28}}{2} = -3 \pm \sqrt{2}$. Thus,the roots are $2+\sqrt{3}, 2-\sqrt{3}, -3+\sqrt{2}, -3-\sqrt{2}$. Comparing with the options,$3-\sqrt{2}$ is not a root.

If the bi-quadratic equation $f(x)=x^4+2x^3-16x^2-22x+7=0$ has $2+\sqrt{3}$ as one of its roots,then which of the following is not a root of $f(x)$?

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