If the roots of the equation (p-3)x^2 + 2(p-3 | vedclass

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(D) Given the quadratic equation $(p-3)x^2 + 2(p-3)x + 2p-5 = 0$. For the roots to be real and distinct,the discriminant $D > 0$.Since the coefficient

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$\frac{4}{5}$

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(D) Given the quadratic equation $(p-3)x^2 + 2(p-3)x + 2p-5 = 0$. For the roots to be real and distinct,the discriminant $D > 0$.Since the coefficient of $x^2$ is $(p-3)$,we must have $p 
eq 3$.$D = [2(p-3)]^2 - 4(p-3)(2p-5) > 0$$4(p-3)^2 - 4(p-3)(2p-5) > 0$Dividing by $4(p-3)$,we get $(p-3) - (2p-5) > 0$ if $p > 3$,or $(p-3) - (2p-5) Case $1$: $p > 3$: $-p + 2 > 0 \Rightarrow p Case $2$: $p  2$.Thus,$2 The expression becomes $f(x) = -(2+3)x^2 + (2 	imes 3)x + (2-3) = -5x^2 + 6x - 1$.The extreme value (maximum) of a quadratic $ax^2 + bx + c$ where $a Here $a = -5, b = 6, c = -1$.Extreme value $= \frac{4(-5)(-1) - (6)^2}{4(-5)} = \frac{20 - 36}{-20} = \frac{-16}{-20} = \frac{4}{5}$.

If the roots of the equation $(p-3)x^2 + 2(p-3)x + 2p-5 = 0$ are real and distinct for $\alpha < p < \beta$ and $(\beta - \alpha)$ is maximum,then the extreme value of the quadratic expression $-(\alpha + \beta)x^2 + \alpha \beta x + (\alpha - \beta)$ is

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