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(C) Comparing $x^{2}-5x+6=0$ with the standard form $ax^{2}+bx+c=0$,we get $a=1, b=-5, c=6$.The sum of the roots is $\alpha+\beta = -b/a = 5/1 = 5$.Th

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$6x^{2}-5x+1=0$

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(C) Comparing $x^{2}-5x+6=0$ with the standard form $ax^{2}+bx+c=0$,we get $a=1, b=-5, c=6$.The sum of the roots is $\alpha+\beta = -b/a = 5/1 = 5$.The product of the roots is $\alpha\beta = c/a = 6/1 = 6$.We need to form a quadratic equation whose roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.The required equation is given by $x^{2} - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.Sum of new roots = $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{5}{6}$.Product of new roots = $\frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha\beta} = \frac{1}{6}$.Substituting these values into the equation: $x^{2} - \frac{5}{6}x + \frac{1}{6} = 0$.Multiplying the entire equation by $6$,we get $6x^{2}-5x+1=0$.

If $\alpha, \beta$ are the roots of the equation $x^{2}-5x+6=0$,construct a quadratic equation whose roots are $\frac{1}{\alpha}, \frac{1}{\beta}$.

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