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(A) Given that $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.Let the n

Vedclass · Accepted Answer

$acx^2 + (a + c)bx + (a + c)^2 = 0$

Vedclass · Answer

(A) Given that $\alpha$ and $\beta$ are roots of $ax^2 + bx + c = 0$,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.Let the new roots be $S_1 = \alpha + \frac{1}{\beta}$ and $S_2 = \beta + \frac{1}{\alpha}$.Sum of the new roots: $S_1 + S_2 = (\alpha + \beta) + (\frac{1}{\alpha} + \frac{1}{\beta}) = (\alpha + \beta) + \frac{\alpha + \beta}{\alpha\beta} = -\frac{b}{a} + \frac{-b/a}{c/a} = -\frac{b}{a} - \frac{b}{c} = -\frac{b(a+c)}{ac}$.Product of the new roots: $S_1 \cdot S_2 = (\alpha + \frac{1}{\beta})(\beta + \frac{1}{\alpha}) = \alpha\beta + 1 + 1 + \frac{1}{\alpha\beta} = \alpha\beta + 2 + \frac{1}{\alpha\beta} = \frac{c}{a} + 2 + \frac{a}{c} = \frac{c^2 + 2ac + a^2}{ac} = \frac{(a+c)^2}{ac}$.The required quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.$x^2 - [-\frac{b(a+c)}{ac}]x + \frac{(a+c)^2}{ac} = 0$.Multiplying by $ac$,we get $acx^2 + b(a+c)x + (a+c)^2 = 0$.

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$,then the equation whose roots are $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$ is:

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