Q. If the roots of the equation $qx^{2} + px + q = 0$ are complex, where $p,q$ are real, then the roots of the equation $x^{2} - 4qx + p^{2} = 0$ are
A) Real and unequal
B) Real and equal
C) Imaginary
D) None of these
Explanation
(a)
The given equations are
$qx^{2} + px + q = 0$ …(i)
and $x^{2} - 4qx + p^{2} = 0$ …(ii)
Since, root of the Eq. (i) are complex, therefore
$p^{2} - 4q^{2} < 0$
Now, discriminant of Eq. (ii) is
$16q^{2} - 4p^{2} = - 4\left( p^{2} - 4q^{2} \right) > 0$
Hence, roots are real and unequal.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.