If 2+sqrt{3} is a root of the equation f(x)=x | vedclass

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

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$3-\sqrt{2}$

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(A) Given that $2+\sqrt{3}$ is a root of $f(x)=x^4+2x^3-16x^2-22x+7=0$. Since the coefficients are rational,the conjugate $2-\sqrt{3}$ must also be a root. Let $x=2+\sqrt{3}$,then $(x-2)^2=3$,which simplifies to $x^2-4x+1=0$. Dividing $f(x)$ by $x^2-4x+1$,we get the quotient $x^2+6x+7$. Setting $x^2+6x+7=0$,we use the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: $x = \frac{-6 \pm \sqrt{36-28}}{2} = \frac{-6 \pm \sqrt{8}}{2} = -3 \pm \sqrt{2}$. The roots of $f(x)=0$ are $2+\sqrt{3}, 2-\sqrt{3}, -3+\sqrt{2}, -3-\sqrt{2}$. Comparing these with the given options,$3-\sqrt{2}$ is not a root.

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

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