Question

The number of real roots of

( x + 1/x )³ + ( x + 1/x ) = 0 is ?

explain

Answer 1

Let x + 1/x = y.
Since the minimum value of y is 2 for a real x, y cannot be 0.

Given y^3 + y = 0
=> y×(y^2 +1) = 0
=> y^2 +1 = 0
=> y = + i or -i

So there are no real roots for the given equation in x.

Let us see why y = x +1/x not equal to 0.
x + 1/x = 0
=> x^2 + 1 = x × 0 = 0
=> x = +i or -i.
So for a real x , x+1/x is not 0.

Answer 2

Hello!
Here is your answer

Let's take that

$x + \frac{1}{x} = y$
then the question changes to
$y {}^{3} + y = 0$
$y(y {}^{2} + 1) = 0$
Taking y as common

If
$y(y {}^{2} + 1) = 0$
then
$y = 0 \\ y {}^{2} + 1 = 0$
Therefore
$y = 0 \\ y = i$
Then the number of real solutions is equal to 1

Answer 1......... But Nooooooo

If x + 1/x = 0
x^2 + 1 = 0
x = ± i
Answer = 0........