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

Let $S_n$ denote the sum of the first $n$ terms of an arithmetic progression. If $S_{10} = 390$ and the ratio of the tenth and the fifth terms is $15:7$,then $S_{15} - S_5$ is equal to:

If the $n$ terms $a_1, a_2, \ldots, a_n$ are in $A$.$P$. with common difference $r$, then the difference between the mean of their squares and the square of their mean is

If the $n^{th}$ term of an arithmetic progression is $\frac{(2n + 1)}{3}$,what is the sum of its first $19$ terms?

The ${n^{th}}$ term of the series $3 \cdot 8 + 6 \cdot 11 + 9 \cdot 14 + 12 \cdot 17 + \dots$ will be

$A$ farmer buys a used tractor for $Rs. 12000$. He pays $Rs. 6000$ cash and agrees to pay the balance in annual instalments of $Rs. 500$ plus $12\%$ interest on the unpaid amount. How much will the tractor cost him?

Vedclass Test Series

Exam Paper Generator

Online Exam Module