Question

Let $p\left(x\right)$ be cubic polynomial $7x$ ${}^3$ $-4x$ ${}^2$ $+K$. Suppose the three roots of $p\left(x\right)$ form an arithmetic progression. Then the value of $K,$ is

• A. $\frac{4}{21}$
• B. $\frac{16}{147}$
• C. $\frac{16}{441}$
• D. $\frac{128}{1323}$

Answer 1

The correct option is D $\frac{128}{1323}$

Given cubic expression $p\left(x\right)=7x^3-4x^2+K$
Comparing with the standard cubic expression $p\left(x\right)=a{x}^3+b{x}^2+cx+d$, we get
$a=7,b=-4,c=0,d=K$
The roots of the equation are in arithmetic progression.

Let them be $m-n,m$ and $m+n.$
The sum of the roots $=\frac{-b}{a}$ = $\frac{-(-4)}{7}$ = $\frac{4}{7}=(m-n)+m+(m-n)=3m$ ....(1)
Sum of bi-product of roots $=\frac{c}{a}$ = $\frac{0}{7}$ = $0$ $=(m-n)m+(m-n)(m+n)+m(m+n)=$ $2m^2+m^2-n^2$ .....(2)
Product of roots $=\frac{-d}{a}$ = $\frac{-K}{7}$ = $(m-n)m(m+n)$ = $m(m^2-n^2)$ ........(3)
From equations (1),(2) and (3), we get
$m=\frac{4}{21}$ ........(4)
$2m^2+m^2-n^2=0$ .....(5)
$m^2-n^2=-\frac{3K}{4}$ .....(6)
Substitute equation (4) and (6) in equation (5), we get
$2{\left(\frac{4}{21}\right)}^2-\frac{3K}{4}=0$
$\Rightarrow$$K=\left(\frac{8}{3}\right){\left(\frac{4}{21}\right)}^2$ = $K=\left(\frac{8}{3}\right)\left(\frac{16}{441}\right)$
$\Rightarrow$$K=\frac{128}{1323}$
Hence, option (D) is the correct answer.