Question

Let ${\alpha }_1,{\alpha }_2$ be the roots of $x^2-x+p=0$ and ${\alpha }_3,{\alpha }_4$ be the roots of $x^2-4x+q=0$. If
${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4$ are in GP, then the integral values of $p$ and $q$ respectively are

• A. $-2,-32$
• B. $-2,3$
• C. $-6,3$
• D. $-6,-32$

Answer 1

The correct option is A $-2,-32$

Given ${\alpha }_1,{\alpha }_2$ are the roots of $x^2-x+p=0$

$\Rightarrow {\alpha }_1+{\alpha }_2=1$ , ${\alpha }_1{\alpha }_2=p$

Also given ${\alpha }_3,{\alpha }_4$ are the roots of $x^2-4x+q=0$

$\Rightarrow {\alpha }_3+{\alpha }_4=4$ , ${\alpha }_3{\alpha }_4=q$

Now given ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_4$ are in G.P

$\Rightarrow \frac{{\alpha }_2}{{\alpha }_1}=\frac{{\alpha }_3}{{\alpha }_2}=\frac{{\alpha }_4}{{\alpha }_3}$

$\Rightarrow \frac{{\alpha }_1}{{\alpha }_2}=\frac{{\alpha }_3}{{\alpha }_4}$ (Taking first and third part and applying invertendo)

Applying componendo, we get

$\frac{{\alpha }_1+{\alpha }_2}{{\alpha }_3+{\alpha }_4}=\frac{{\alpha }_2}{{\alpha }_4}$

Squaring both sides

$\frac{({\alpha }_1+{\alpha }_2{)}^2}{({\alpha }_3+{\alpha }_4{)}^2}=\frac{{\alpha }_2}{{\alpha }_4}\frac{{\alpha }_2}{{\alpha }_4}$

$\frac{({\alpha }_1+{\alpha }_2{)}^2}{({\alpha }_3+{\alpha }_4{)}^2}=\frac{{\alpha }_1}{{\alpha }_3}\frac{{\alpha }_2}{{\alpha }_4}$

$\frac{1}{16}=\frac{p}{q}$

Option A, satisfies the above ratio.

$\Rightarrow p=-2,q=-32$