If alpha, beta are the roots of the equation | vedclass

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Question

(A) Given $\alpha + \beta = -\frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$.Let the new roots be $S_1 = \alpha + \frac{1}{\beta}$ and $S_2 = \beta + \f

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given $\alpha + \beta = -\frac{b}{a}$ and $\alpha \beta = \frac{c}{a}$.Let the new roots be $S_1 = \alpha + \frac{1}{\beta}$ and $S_2 = \beta + \frac{1}{\alpha}$.The sum of the new roots is $S_1 + S_2 = (\alpha + \beta) + \frac{\alpha + \beta}{\alpha \beta} = -\frac{b}{a} + \frac{-b/a}{c/a} = -\frac{b}{a} - \frac{b}{c} = -\frac{b(a+c)}{ac}$.The product of the new roots is $S_1 S_2 = (\alpha + \frac{1}{\beta})(\beta + \frac{1}{\alpha}) = \alpha \beta + 1 + 1 + \frac{1}{\alpha \beta} = \alpha \beta + 2 + \frac{1}{\alpha \beta} = \frac{c}{a} + 2 + \frac{a}{c} = \frac{c^2 + 2ac + a^2}{ac} = \frac{(a+c)^2}{ac}$.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 $acx^2 + b(a+c)x + (a+c)^2 = 0$.

List-$I$	List-$II$
$(i) \alpha = \beta$	$(A) (ac^2)^{1/3} + (a^2c)^{1/3} + b = 0$
$(ii) \alpha = 2\beta$	$(B) 2b^2 = 9ac$
$(iii) \alpha = 3\beta$	$(C) b^2 = 6ac$
$(iv) \alpha = \beta^2$	$(D) 3b^2 = 16ac$
	$(E) b^2 = 4ac$
	$(F) (ac^2)^{1/3} + (a^2c)^{1/3} = b$

If $\alpha, \beta$ 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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