If the roots of the quadratic equation ax^2+b | vedclass

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

Question

(C) Given that the quadratic equation $ax^2+bx+c=0$ has imaginary roots,the discriminant $D = b^2 - 4ac < 0$,which implies $b^2 < 4ac$.Let $f(x) = 3a^

Vedclass · Accepted Answer

$> -4ac$

Vedclass · Answer

(C) Given that the quadratic equation $ax^2+bx+c=0$ has imaginary roots,the discriminant $D = b^2 - 4ac Let $f(x) = 3a^2x^2 + 6abx + 2b^2$.Since the coefficient of $x^2$ is $3a^2 > 0$,the expression has a minimum value.The minimum value of a quadratic $Ax^2 + Bx + C$ is given by $\frac{4AC - B^2}{4A}$.Here,$A = 3a^2$,$B = 6ab$,and $C = 2b^2$.Minimum value $= \frac{4(3a^2)(2b^2) - (6ab)^2}{4(3a^2)} = \frac{24a^2b^2 - 36a^2b^2}{12a^2} = \frac{-12a^2b^2}{12a^2} = -b^2$.Since $b^2  -4ac$.Thus,the minimum value is greater than $-4ac$.

If the roots of the quadratic equation $ax^2+bx+c=0$ are imaginary,then for all real values of $x$,the minimum value of the expression $3a^2x^2+6abx+2b^2$ is

Similar Questions

If one root of the equation $x^3-6x^2+3x+10=0$ is the average of the other two,then the sum of the fourth powers of the roots of the equation is

Find the product of all real roots of the equation $|x|^{6/5} - 26|x|^{3/5} - 27 = 0$.

Let $f: R \rightarrow R$ be the function $f(x) = (x - a_1)(x - a_2) + (x - a_2)(x - a_3) + (x - a_3)(x - a_1)$ with $a_1, a_2, a_3 \in R$. Then,$f(x) \geq 0$ if and only if

Suppose that $x$ and $y$ are positive numbers with $xy = \frac{1}{9}$,$x(y + 1) = \frac{7}{9}$,and $y(x + 1) = \frac{5}{18}$. The value of $(x + 1)(y + 1)$ is equal to:

If one root of the equation $x^2 + px + 12 = 0$ is $4$,and the equation $x^2 + px + q = 0$ has equal roots,then what is the value of $q$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module