If a, b, c are real numbers and a 0,what is | vedclass

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

Question

(B) The given quadratic expression is $f(x) = ax^2 + bx + c$ with $a > 0$.To find the minimum value,we complete the square:$f(x) = a(x^2 + \frac{b}{a}

Vedclass · Accepted Answer

$\frac{4ac - b^2}{4a}$

Vedclass · Answer

(B) The given quadratic expression is $f(x) = ax^2 + bx + c$ with $a > 0$.To find the minimum value,we complete the square:$f(x) = a(x^2 + \frac{b}{a}x) + c$$f(x) = a(x^2 + \frac{b}{a}x + (\frac{b}{2a})^2 - (\frac{b}{2a})^2) + c$$f(x) = a(x + \frac{b}{2a})^2 - \frac{b^2}{4a} + c$$f(x) = a(x + \frac{b}{2a})^2 + \frac{4ac - b^2}{4a}$Since $a > 0$,the term $a(x + \frac{b}{2a})^2 \ge 0$ for all real $x$.The minimum value occurs when $x = -\frac{b}{2a}$,and the minimum value is $\frac{4ac - b^2}{4a}$.

If $a, b, c$ are real numbers and $a > 0$,what is the minimum value of the quadratic expression $ax^2 + bx + c$ for real $x$?

Similar Questions

The value of $k$ for which the equation $x^2 - 3x + k = 0$ has at least one real root in $[0, 1]$ is

The quadratic expression $(2x+1)^{2} - px + q \neq 0$ for any real $x$, if

For a real number $x$,if the minimum value of $f(x) = x^2 + 2bx + 2c^2$ is greater than the maximum value of $g(x) = -x^2 - 2cx + b^2$,then:

If exactly one root of the equation $x^2 + (a - 1)x + 2a = 0$ lies in the interval $(0, 3)$,then the set of values of $a$ is given by:

Number of integral values of $a$ for which both roots of the quadratic equation $x^2 - (2a + 3)x + a^2 + 3a = 0$ lie in the interval $(0, 4)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module