Real roots

$1+x+x^2+x^3 = x^4 + x^5\\ 1(x+1) + x^2(x+1) = x^4(x+1)\\ (x^4 - x^2 - 1)(x+1) = 0\\ x=-1 \quad x^4-x^2-1=0$

Here, we found one real root.

Next, taking the second factor:

$x^4 - x^2 -1 = 0\\ x^2 = \dfrac{1 \pm \sqrt{(-1)^2 - 4(1)(-1)}}{2(1)}\\ = \dfrac{1 \pm \sqrt{1+4}}{2}\\ = \dfrac{1 \pm \sqrt{5}}{2}$

When $x^2 = \dfrac{1 + \sqrt{5}}{2}$

$x = \pm \sqrt{\dfrac{1 + \sqrt{5}}{2}}$

That's two more real roots.

When $x^2 = \dfrac{1 - \sqrt{5}}{2}$

$x = \pm \sqrt{\dfrac{1 - \sqrt{5}}{2}}$

Note that $\dfrac{1 - \sqrt{5}}{2}$ is a negative number.

This means that these last two roots are complex roots.

Therefore, we have $\boxed{3}$ real roots for this equation

Factorizing both sides,

(X+1)(X^2 + 1) = X^4 (X + 1)...

→(x+1)(x^4 - x^2 -1)=0...

Which gives one root x = -1, & for other solutions,

Set x^2 = t (t > 0) , to get,

t^2 - t - 1= 0

Solving this we get ,

t = (1+root5)/2 &

t = (1-root5)/2 , which we will neglect bcoz, t > 0......

Hence putting +ve value of t in x^2 ...

We get two other values of x...

Hence total no. of real solutions of x are 3...

Thank you!