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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

The electrostatic potential inside a charged spherical ball is given by $\phi = ar^2 + b$,where $r$ is the distance from the centre and $a, b$ are constants. Then the charge density inside the ball is

The major product of the following reaction is

$A$ conducting ring is placed around the core of an electromagnet as shown in the figure. When key $K$ is pressed,the ring

Which of the following is $NOT$ a method of vegetative reproduction in Bryophytes?

Vedclass Test Series

Exam Paper Generator

Online Exam Module