2+sqrt{5} and 1 are roots of the cubic equati | vedclass

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Question

(C) We know that a cubic equation with rational coefficients having an irrational root $\alpha+\sqrt{\beta}$ (where $\alpha, \beta \in \mathbb{Q}$ and

Vedclass · Accepted Answer

$x^3-5x^2+3x+1=0$

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(C) We know that a cubic equation with rational coefficients having an irrational root $\alpha+\sqrt{\beta}$ (where $\alpha, \beta \in \mathbb{Q}$ and $\beta$ is not a perfect square) must also have the conjugate root $\alpha-\sqrt{\beta}$.Hence,the roots are $1, 2+\sqrt{5},$ and $2-\sqrt{5}$.The cubic equation with roots $\alpha, \beta, \gamma$ is given by $x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-(\alpha\beta\gamma)=0$.Let $\alpha=1, \beta=2+\sqrt{5}, \gamma=2-\sqrt{5}$.Sum of roots: $\alpha+\beta+\gamma = 1+(2+\sqrt{5})+(2-\sqrt{5}) = 5$.Sum of product of roots taken two at a time: $\alpha\beta+\beta\gamma+\gamma\alpha = 1(2+\sqrt{5})+(2+\sqrt{5})(2-\sqrt{5})+1(2-\sqrt{5}) = 2+\sqrt{5}+4-5+2-\sqrt{5} = 3$.Product of roots: $\alpha\beta\gamma = 1(2+\sqrt{5})(2-\sqrt{5}) = 1(4-5) = -1$.Substituting these into the general form: $x^3-(5)x^2+(3)x-(-1) = 0$,which simplifies to $x^3-5x^2+3x+1=0$.

$2+\sqrt{5}$ and $1$ are roots of the cubic equation given by

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