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(A) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{

Vedclass · Accepted Answer

$\frac{1}{\alpha}, \frac{1}{\beta}$

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(A) Given the quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$,we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.Consider the equation $cx^2 + bx + a = 0$. Let its roots be $x_1$ and $x_2$.For this equation,the sum of roots is $x_1 + x_2 = -\frac{b}{c}$ and the product of roots is $x_1x_2 = \frac{a}{c}$.If we substitute $x = \frac{1}{y}$ into the original equation $ax^2 + bx + c = 0$,we get $a(\frac{1}{y})^2 + b(\frac{1}{y}) + c = 0$.Multiplying by $y^2$,we get $a + by + cy^2 = 0$,which is $cy^2 + by + a = 0$.Since $x = \frac{1}{y}$,the roots of the new equation are the reciprocals of the roots of the original equation.Therefore,the roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$.

If $\alpha$ and $\beta$ are the roots of the equation $ax^2 + bx + c = 0$,then what are the roots of the equation $cx^2 + bx + a = 0$?

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