The equation whose roots are frac{1}{3 + sqrt | vedclass

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(A) Let the roots be $\alpha = \frac{1}{3 + \sqrt{2}}$ and $\beta = \frac{1}{3 - \sqrt{2}}$.Rationalizing the roots:$\alpha = \frac{3 - \sqrt{2}}{(3 +

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$7x^2 - 6x + 1 = 0$

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(A) Let the roots be $\alpha = \frac{1}{3 + \sqrt{2}}$ and $\beta = \frac{1}{3 - \sqrt{2}}$.Rationalizing the roots:$\alpha = \frac{3 - \sqrt{2}}{(3 + \sqrt{2})(3 - \sqrt{2})} = \frac{3 - \sqrt{2}}{9 - 2} = \frac{3 - \sqrt{2}}{7}$$\beta = \frac{3 + \sqrt{2}}{(3 - \sqrt{2})(3 + \sqrt{2})} = \frac{3 + \sqrt{2}}{9 - 2} = \frac{3 + \sqrt{2}}{7}$Sum of roots $(\alpha + \beta) = \frac{3 - \sqrt{2} + 3 + \sqrt{2}}{7} = \frac{6}{7}$Product of roots $(\alpha \beta) = \frac{(3 - \sqrt{2})(3 + \sqrt{2})}{7 	imes 7} = \frac{9 - 2}{49} = \frac{7}{49} = \frac{1}{7}$The quadratic equation is given by $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.$x^2 - \frac{6}{7}x + \frac{1}{7} = 0$Multiplying by $7$,we get $7x^2 - 6x + 1 = 0$.

The equation whose roots are $\frac{1}{3 + \sqrt{2}}$ and $\frac{1}{3 - \sqrt{2}}$ is

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