Question

Let ${\alpha }_1,{\alpha }_2,$ be the root of $x^2-4x+k_1=0,{\alpha }_3,{\alpha }_4$ be the root of $x^2-36x+k_2=0$ where ${\alpha }_1<{\alpha }_2<{\alpha }_3<{\alpha }_4$ are in G.P ., then the value of $k_1{k}_2$ is

• A. $81$
• B. $243$
• C. $729$
• D. $27$

Answer 1

Answer: (C)

$729$

As ${\alpha }_1,{\alpha }_2$ are roots of $x^2-4x+k_1=0$
We have
${\alpha }_1+{\alpha }_2=4$ ...(1)
${\alpha }_1{\alpha }_2=k_1$ ...(2)
And as ${\alpha }_3,{\alpha }_4$ are roots of $x^2-36x+k_2=0$
We have
${\alpha }_3+{\alpha }_4=36$ ...(3)
${\alpha }_3{\alpha }_4=k_2$ ...(4)
Now as ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4$ are in G.P
Let $r$ is the common ratio then
From (1) ${\alpha }_1+{\alpha }_2={\alpha }_1(1+r)=4$ ...(5)
And from (3) ${\alpha }_3+{\alpha }_4={\alpha }_1{r}^2(1+r)=36$ ...(6)
Dividing (6) by (5), we get
$r^2=9\Rightarrow r=\pm 3r=3\Rightarrow {\alpha }_1=1,{\alpha }_2=3,{\alpha }_3=9,{\alpha }_4=27$
Hence from this and from (2) and (4)
$k_1{k}_2={\alpha }_1{\alpha }_2{\alpha }_3{\alpha }_4=729$