If the roots of the equation {x^3} - 12{x^2} | vedclass

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

Question

(C) Let the roots of the cubic equation ${x^3} - 12{x^2} + 39x - 28 = 0$ be $a - d, a, a + d$,which are in $A.P.$According to the properties of roots

Vedclass · Accepted Answer

$\pm 3$

Vedclass · Answer

(C) Let the roots of the cubic equation ${x^3} - 12{x^2} + 39x - 28 = 0$ be $a - d, a, a + d$,which are in $A.P.$According to the properties of roots of a cubic equation:Sum of roots = $(a - d) + a + (a + d) = -(-12)/1 = 12$$3a = 12 \implies a = 4$Product of roots = $(a - d) \cdot a \cdot (a + d) = -(-28)/1 = 28$$a(a^2 - d^2) = 28$Substitute $a = 4$ into the product equation:$4(4^2 - d^2) = 28$$16 - d^2 = 7$$d^2 = 9$$d = \pm 3$Thus,the common difference is $\pm 3$.

If the roots of the equation ${x^3} - 12{x^2} + 39x - 28 = 0$ are in $A.P.$,then their common difference will be

Similar Questions

Find the sum of the numbers of three digits divisible by $7$.

If ${S_n} = nP + \frac{1}{2}n(n - 1)Q$,where ${S_n}$ denotes the sum of the first $n$ terms of an $A.P.$,then the common difference is

If $|\alpha| < 1$ and $|\beta| < 1$,where $s_1 = 1 - \alpha + \alpha^2 - \alpha^3 + \dots \infty$ and $s_2 = 1 - \beta + \beta^2 - \beta^3 + \dots \infty$,then $1 - \alpha\beta + \alpha^2\beta^2 - \alpha^3\beta^3 + \dots \infty$ equals:

If $1, \log_y x, \log_z y, -15 \log_x z$ are in $A.P.$,then

If three positive numbers $a, b$ and $c$ are in $A.P.$ such that $abc = 8$,then the minimum possible value of $b$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module