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 can complete

Vedclass · Accepted Answer

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

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 can complete the square:$y = x^2 - (\alpha+\beta)x + \left(\frac{\alpha+\beta}{2}\right)^2 - \left(\frac{\alpha+\beta}{2}\right)^2 + \alpha\beta$.$y = \left(x - \frac{\alpha+\beta}{2}\right)^2 - \frac{(\alpha+\beta)^2 - 4\alpha\beta}{4}$.$y = \left(x - \frac{\alpha+\beta}{2}\right)^2 - \frac{\alpha^2 + 2\alpha\beta + \beta^2 - 4\alpha\beta}{4}$.$y = \left(x - \frac{\alpha+\beta}{2}\right)^2 - \frac{(\alpha-\beta)^2}{4}$.Since the square term $\left(x - \frac{\alpha+\beta}{2}\right)^2 \ge 0$,the minimum value occurs when $x = \frac{\alpha+\beta}{2}$.Thus,the minimum value is $-\frac{1}{4}(\alpha-\beta)^2$.

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

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