If alpha and beta are the roots of the equati | vedclass

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(B) Given $\alpha, \beta$ are roots of $x^2 - 2x + 3 = 0$. From the relation between roots and coefficients,$\alpha + \beta = 2$ and $\alpha \beta = 3

Vedclass · Accepted Answer

$9x^2 + 2x + 1 = 0$

Vedclass · Answer

(B) Given $\alpha, \beta$ are roots of $x^2 - 2x + 3 = 0$. From the relation between roots and coefficients,$\alpha + \beta = 2$ and $\alpha \beta = 3$. We need to find the equation with roots $\frac{1}{\alpha^2}$ and $\frac{1}{\beta^2}$. The sum of the new roots is $\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha \beta)^2} = \frac{(\alpha + \beta)^2 - 2\alpha \beta}{(\alpha \beta)^2} = \frac{2^2 - 2(3)}{3^2} = \frac{4 - 6}{9} = -\frac{2}{9}$. The product of the new roots is $\frac{1}{\alpha^2} \cdot \frac{1}{\beta^2} = \frac{1}{(\alpha \beta)^2} = \frac{1}{3^2} = \frac{1}{9}$. The required quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$. $x^2 - (-\frac{2}{9})x + \frac{1}{9} = 0$. Multiplying by $9$,we get $9x^2 + 2x + 1 = 0$.

If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 2x + 3 = 0$,then the equation whose roots are $\frac{1}{\alpha^2}$ and $\frac{1}{\beta^2}$ is

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