Search any question & find its solution
Question:
Answered & Verified by Expert
One root ofthe equation $x^{2}=p x+q$ is reciprocal of the othe and $\mathrm{p} \neq \pm 1$. What is the value of q?
Options:
Solution:
1585 Upvotes
Verified Answer
The correct answer is:
$q=-1$
Let the roots of the equation $x^{2}-p x-q=0$ be and $\frac{1}{\alpha}$
$\Rightarrow$ Product of roots $=\alpha \cdot \frac{1}{\alpha}=-\frac{\mathrm{q}}{1}$
$1=-\mathrm{q} \Rightarrow \mathrm{q}=-1$
$\Rightarrow$ Product of roots $=\alpha \cdot \frac{1}{\alpha}=-\frac{\mathrm{q}}{1}$
$1=-\mathrm{q} \Rightarrow \mathrm{q}=-1$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.