Question

Consider the equation $3 x^4-18 x^3+p x^2-8 q x+3 q=0$. The equation has only positive real roots then the value of $\frac{p}{q}$ is

• A. $\frac{1}{8}$
• B. 4
• C. $\frac{1}{4}$
• D. 8

Answer 1

Answer: (D)

$\text { A.M. }=\frac{x_1+x_2+x_3+x_4}{4}=\frac{18}{3 \times 4}=\frac{3}{2}$
$\text { H.M. }=\frac{4}{8 / 3}=\frac{12}{8}=\frac{3}{2} $
$\text { A.M. }=\text { H.M. } \Rightarrow x_1=x_2=x_3=x_4=\frac{3}{2}$
$p =\frac{81}{2}$ and $q =\frac{81}{16}$.