The quadratic equation whose one root is frac | vedclass

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Question

(A) Given one root $\alpha = \frac{1}{2 + \sqrt{5}}$.Rationalizing the denominator,we get $\alpha = \frac{2 - \sqrt{5}}{(2 + \sqrt{5})(2 - \sqrt{5})}

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$x^2 + 4x - 1 = 0$

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(A) Given one root $\alpha = \frac{1}{2 + \sqrt{5}}$.Rationalizing the denominator,we get $\alpha = \frac{2 - \sqrt{5}}{(2 + \sqrt{5})(2 - \sqrt{5})} = \frac{2 - \sqrt{5}}{4 - 5} = \frac{2 - \sqrt{5}}{-1} = \sqrt{5} - 2$.Since the coefficients of the quadratic equation are typically rational,the other root $\beta$ must be the conjugate of $\alpha$,so $\beta = -\sqrt{5} - 2$.Sum of roots $(\alpha + \beta) = (\sqrt{5} - 2) + (-\sqrt{5} - 2) = -4$.Product of roots $(\alpha \beta) = (\sqrt{5} - 2)(-\sqrt{5} - 2) = -(\sqrt{5} - 2)(\sqrt{5} + 2) = -(5 - 4) = -1$.The quadratic equation is given by $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Substituting the values,we get $x^2 - (-4)x + (-1) = 0$,which simplifies to $x^2 + 4x - 1 = 0$.

The quadratic equation whose one root is $\frac{1}{2 + \sqrt{5}}$ will be

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