If the roots of the equation x^3+ax^2+bx+c=0 | vedclass

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

Question

(B) Let the roots of the cubic equation be $A-d, A, A+d$. Since the sum of the roots is $-a$,we have $(A-d) + A + (A+d) = -a$,which gives $3A = -a$,so

Vedclass · Accepted Answer

$9ab=2a^3+27c$

Vedclass · Answer

(B) Let the roots of the cubic equation be $A-d, A, A+d$. Since the sum of the roots is $-a$,we have $(A-d) + A + (A+d) = -a$,which gives $3A = -a$,so $A = -\frac{a}{3}$. Since $A$ is a root,it must satisfy the equation: $A^3 + aA^2 + bA + c = 0$. Substituting $A = -\frac{a}{3}$: $(-\frac{a}{3})^3 + a(-\frac{a}{3})^2 + b(-\frac{a}{3}) + c = 0$ $-\frac{a^3}{27} + \frac{a^3}{9} - \frac{ab}{3} + c = 0$ Multiplying by $27$: $-a^3 + 3a^3 - 9ab + 27c = 0$ $2a^3 - 9ab + 27c = 0$ $9ab = 2a^3 + 27c$.

If the roots of the equation $x^3+ax^2+bx+c=0$ are in arithmetic progression,then

Similar Questions

If the sum of the first $11$ terms of an $A.P.$,$a_{1}, a_{2}, a_{3}, \ldots$ is $0$ $(a_{1} \neq 0)$,then the sum of the $A.P.$,$a_{1}, a_{3}, a_{5}, \ldots, a_{23}$ is $k a_{1}$,where $k$ is equal to

The four arithmetic means between $3$ and $23$ are.....

The sum of all natural numbers $n$ such that $100 < n < 200$ and $H.C.F. (91, n) > 1$ is

If the sum of the first $n$ even natural numbers is $k$ times the sum of the first $n$ odd natural numbers,then $k = ........$

If $1, \log_9(3^{1-x} + 2), \log_3(4 \cdot 3^x - 1)$ are in an arithmetic progression,then $x = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module