Question

If $p,q,r$ are the roots of the equation $\left|\begin{array}{ccc}x& 1& 2\\ 1& x& 2\\ 1& 2& x\end{array}\right|=0$ , then $\frac{p^4+q^4+r^4}{p^2+q^2+r^2}$ is equal to

• A. 7
• B. 6
• C. 1/6
• D. 1/7

Answer 1

Answer: (A)

$p,q,r$ the roots of equation ,
$\left|\begin{array}{ccc}x& 1& 2\\ 1& x& 2\\ 1& 2& x\end{array}\right|=0$
Operating $C_1\to C_1+C_2+C_3$, we get
$\left|\begin{array}{ccc}x+3& 1& 2\\ x+3& x& 2\\ x+3& 2& x\end{array}\right|=0$
$\Rightarrow \left(x+3\right)\left|\begin{array}{ccc}1& 1& 2\\ 1& x& 2\\ 1& 2& x\end{array}\right|=0$
Operating $R_2\to R_2-R_1,R_3\to R_3-R_1$
$\Rightarrow \left(x+3\right)\left|\begin{array}{ccc}1& 1& 2\\ 0& x-1& 0\\ 0& 1& x-2\end{array}\right|$
Expanding along $C_1$, we get $(x+3)[(x-l)(x-2)-0]=0x=1,2,-3$ i.e., roots of equation
Let $p=1,q=2,r=-3$.
$\therefore \frac{p^4+q^4+r^4}{p^2+q^2+r^2}=\frac{1+16+81}{1+4+9}=\frac{98}{14}=7$