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

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(B) Given the quadratic equation $2x^2 - (p + 1)x + (p - 1) = 0$.Comparing with $ax^2 + bx + c = 0$,we have $a = 2$,$b = -(p + 1)$,and $c = (p - 1)$.S

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$2$

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(B) Given the quadratic equation $2x^2 - (p + 1)x + (p - 1) = 0$.Comparing with $ax^2 + bx + c = 0$,we have $a = 2$,$b = -(p + 1)$,and $c = (p - 1)$.Sum of roots: $\alpha + \beta = -b/a = (p + 1)/2$.Product of roots: $\alpha \beta = c/a = (p - 1)/2$.Given the condition $\alpha - \beta = \alpha \beta$.Squaring both sides: $(\alpha - \beta)^2 = (\alpha \beta)^2$.Using the identity $(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha \beta$,we get:$(\alpha + \beta)^2 - 4\alpha \beta = (\alpha \beta)^2$.Substituting the values:$((p + 1)/2)^2 - 4((p - 1)/2) = ((p - 1)/2)^2$.$(p + 1)^2/4 - 2(p - 1) = (p - 1)^2/4$.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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