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(D) Let the roots of the quadratic equation $2x^2 - (a + 1)x + (a - 1) = 0$ be $\alpha$ and $\beta$.From the properties of roots,the sum of roots $\al

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$2$

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(D) Let the roots of the quadratic equation $2x^2 - (a + 1)x + (a - 1) = 0$ be $\alpha$ and $\beta$.From the properties of roots,the sum of roots $\alpha + \beta = \frac{a + 1}{2}$ and the product of roots $\alpha \beta = \frac{a - 1}{2}$.The difference of the roots is given by $|\alpha - \beta| = \sqrt{(\alpha + \beta)^2 - 4\alpha \beta}$.Given that the difference of the roots is equal to their product,we have $|\alpha - \beta| = \alpha \beta$.Squaring both sides,we get $(\alpha - \beta)^2 = (\alpha \beta)^2$.Substituting the values,$(\alpha + \beta)^2 - 4\alpha \beta = (\alpha \beta)^2$.$(\frac{a + 1}{2})^2 - 4(\frac{a - 1}{2}) = (\frac{a - 1}{2})^2$.$\frac{a^2 + 2a + 1}{4} - 2(a - 1) = \frac{a^2 - 2a + 1}{4}$.Multiplying by $4$,$a^2 + 2a + 1 - 8(a - 1) = a^2 - 2a + 1$.$a^2 + 2a + 1 - 8a + 8 = a^2 - 2a + 1$.$-6a + 9 = -2a + 1$.$8 = 4a$.$a = 2$.

For what value of $a$ is the difference of the roots of the equation $2x^2 - (a + 1)x + (a - 1) = 0$ equal to their product?

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