Question

Let $\alpha, \beta$ be the roots of $x^2 - x + p = 0$ and $y, \delta$ be the roots of $x^2 - 4x + q = 0$ if $\alpha, \beta, y \delta$ are in GP, then the integer values of p and q respectively are

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

Answer 1

Answer: (A)

-2,-32

Answer 2

\begin{array} \ \alpha + \beta = 1 \\\ \ \ \ \ \ \alpha \beta = p \\\ \end{array} \Bigg \\} \ and \ \ \ \begin{array} \ \lambda + \delta = 4 \\\ \ \ \ \ \ \lambda \delta = q \\\ \end{array} \Bigg \\}
Let r be the common ratio.
Since, $\alpha \beta, y\ and\ \delta$ are in GP.
Therefore
\hspace35mm \beta = \alpha r, y= \alpha r^2
and \hspace25mm \delta = \alpha r^3
Then,\hspace15mm \alpha +\alpha r = 1 \Rightarrow \alpha (1+r) = 1 \hspace20mm ...(i)
and \hspace15mm \alpha r^2 +\alpha r^3 = 4 \Rightarrow \alpha r^2(1+r) = 4 \hspace15mm ...(ii)
From Eqs. (i) and (ii), $r^2 = 4 \Rightarrow r = \pm 2$
Now, \hspace20mm \alpha . \alpha r = p\ and $\alpha r^2 .\alpha r^3 = q$
On putting \hspace15mm r = - 2 we get
\hspace35mm \alpha = - 1 ,p = - 2 and $q = - 32$
Again putting r = 2, we get $\alpha = 1/3\$ and $p = - \frac{2}{9}$
Since, q and p are integers.
Therefore, we take p = - 2 and q = - 32.