JEE Advanced | 2014Quadratic EquationHard
Question
The quadratic equation p(x) = 0 with real coefficients has purely imaginary roots. Then the equation p(p(x)) = 0 has
Options
A.Only purely imaginary roots
B.All real roots
C.Two real and two purely imaginary roots
D.Neither real nor purely imaginary roots
Solution
Let p(x) = x2 + a (a > 0) (∵ roots are purely imaginary)
p(p(x)) = (x2 + a)2 + a (a ∈ R)
x4 + 2a(x2) + a2 + a = 0
⇒ x2 =
= - a +
x = +
= x1 + iy1
p(p(x)) = (x2 + a)2 + a (a ∈ R)
x4 + 2a(x2) + a2 + a = 0
⇒ x2 =

= - a +

x = +
= x1 + iy1Create a free account to view solution
View Solution FreeMore Quadratic Equation Questions
If α and β be the roots of the equation 2x2 + 2 (a + b) x + a2 + b2 = 0, then the equation whose roots are (&#...If $a,b$ are the roots of the quadratic equation $x^{2} + \lambda x - \frac{1}{2\lambda^{2}} = 0$, where $\lambda$ is a ...The equation whose roots are is-...The number of real solutions of the equation = 10 is -...The possible value(s) of ' $p$ ' for which the equations $ax^{2} - px + ab = 0$ and $x^{2} - ax - bx + ab = 0$ may have ...