The minimum value of (x-alpha)(x-beta) is | vedclass

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(D) Let $y = (x-\alpha)(x-\beta)$.Expanding the expression,we get $y = x^2 - (\alpha+\beta)x + \alpha\beta$.To find the minimum value,we find the deri

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$-\frac{1}{4}(\alpha-\beta)^2$

Vedclass · Answer

(D) Let $y = (x-\alpha)(x-\beta)$.Expanding the expression,we get $y = x^2 - (\alpha+\beta)x + \alpha\beta$.To find the minimum value,we find the derivative with respect to $x$ and set it to $0$:$\frac{dy}{dx} = 2x - (\alpha+\beta) = 0$.This gives $x = \frac{\alpha+\beta}{2}$.Since the second derivative $\frac{d^2y}{dx^2} = 2 > 0$,the function has a minimum at $x = \frac{\alpha+\beta}{2}$.Substituting this value into the original expression:$y_{min} = \left(\frac{\alpha+\beta}{2} - \alpha\right)\left(\frac{\alpha+\beta}{2} - \beta\right)$$y_{min} = \left(\frac{\beta-\alpha}{2}\right)\left(\frac{\alpha-\beta}{2}\right)$$y_{min} = -\frac{1}{4}(\alpha-\beta)(\alpha-\beta) = -\frac{1}{4}(\alpha-\beta)^2$.

The minimum value of $(x-\alpha)(x-\beta)$ is

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