For the quadratic equation 2x^2 - (p + 1)x + | vedclass

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(B) Given the quadratic equation $2x^2 - (p + 1)x + (p - 1) = 0$.Let $\alpha$ and $\beta$ be the roots of the equation.From the relationship between r

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$2$

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(B) Given the quadratic equation $2x^2 - (p + 1)x + (p - 1) = 0$.Let $\alpha$ and $\beta$ be the roots of the equation.From the relationship between roots and coefficients:Sum of roots: $\alpha + \beta = \frac{p + 1}{2}$Product of roots: $\alpha \beta = \frac{p - 1}{2}$Given the condition: $\alpha - \beta = \alpha \beta$.Squaring both sides: $(\alpha - \beta)^2 = (\alpha \beta)^2$.We know that $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$.Substituting the values: $(\frac{p + 1}{2})^2 - 4(\frac{p - 1}{2}) = (\frac{p - 1}{2})^2$.$\frac{(p + 1)^2}{4} - 2(p - 1) = \frac{(p - 1)^2}{4}$.Multiply by $4$: $(p + 1)^2 - 8(p - 1) = (p - 1)^2$.$p^2 + 2p + 1 - 8p + 8 = p^2 - 2p + 1$.$-6p + 9 = -2p + 1$.$8 = 4p$.$p = 2$.

For the quadratic equation $2x^2 - (p + 1)x + (p - 1) = 0$,if $\alpha - \beta = \alpha \beta$,then what is the value of $p$?

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