Question

$x_1$ and $x_3$ are the roots of the equation $A{x}^2-4x+1=0$, $x_2$ and $x_4$ are the roots of the equation $B{x}^2-6x+1=0$. lf $x_1,x_2,x_3,x_4$ form a H.P. then $(A,B)=$

• A. $(3,3)$
• B. $(8,8)$
• C. $(3,8)$
• D. $(8,3)$

Answer 1

Answer: (C)

$(3,8)$

$x_1,x_2,x_3,x_4$ are in H.P

$\Rightarrow \frac{1}{x_3}-\frac{1}{x_1}=\frac{1}{x_4}-\frac{1}{x_2}$

$\Rightarrow \frac{x_1-x_3}{x_1{x}_3}=\frac{x_2-x_4}{x_2{x}_4}$

$x_1+x_3=\frac{-b}{a}=\frac{4}{A};x_1{x}_3=\frac{c}{a}=\frac{1}{A.}$

$x_2+x_4=\frac{6}{B};x_2{x}_4=\frac{1}{B}$

$x_1-x_3=\sqrt{(x_1+x_3{)}^2-4x_1{x}_3}=\sqrt{\frac{16}{A^2}-\frac{4}{A}}$

$=\frac{\sqrt{16-4A}}{A}$

$x_2-x_4=\frac{\sqrt{36-4B}}{B}$

$\therefore \frac{x_1-x_3}{x_1{x}_3}=\frac{x_2-x_4}{x_2{x}_4}\Rightarrow \sqrt{16-4A}=\sqrt{36-4B}$

$\Rightarrow (A,B)=(3,8)$