JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Quadratic Equations and Inequalities

The quadratic equation

p (x) = 0

with real coefficients has purely imaginary roots. Then the equation

p (p (x)) = 0

has

Answer: (D)

Solution

Since

p (x)

has purely imaginary roots with real coefficients, we can write

p (x) = x^{2} + K

where

K > 0

.

Then:

p (p (x)) = (p (x))^{2} + K = (x^{2} + K)^{2} + K = x^{4} + 2 K x^{2} + K^{2} + K

Let

y = x^{2}

. Then

y^{2} + 2 K y + K^{2} + K = 0

.

Using the quadratic formula:

y = \frac{- 2 K \pm 4 K ^{2} - 4 ( K ^{2} + K )}{2} = \frac{- 2 K \pm - 4 K}{2} = - K \pm i K

Since

y = x^{2}

and

y

is complex (neither real nor purely imaginary),

x

will also be complex but neither real nor purely imaginary.

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option