$x^{4} = 4x + 1$

$x^{4} + 2x^{2} + 1 = 2x^{2} + 4x + 2$

$(x^{2} + 1)^{2} = 2(x + 1)^{2}$

$(x^{2} + 1)^{2} - (\sqrt{2}(x+1))^{2} = 0$

$(x^{2} + 1 +\sqrt{2}(x + 1))((x^{2} + 1 -\sqrt{2}(x + 1)) = 0$

Let us assume that $f(x) = x^{4} - 4x -1$ Differentiating wrt x we get :- $f'(x) = 4x^{3} -4$ We see that $f'(x)$ is $+ve$ for $x>1$ and $-ve$ for x<1.

It is also seen that $f(1) = -4$ i.e. a $-ve$ number.

We see that $f(x)$ keeps on increase after $x=1.$ Thus for $x>1$ it cuts x- axis only once .

Also $f(0) = -1$ i.e. a $-ve$ number.

But $f(x)$ is a decreasing function for $x<1 .$

So $x$ intersects x axis only once before $x<1$ and from the value of $f(0)$ it is evident that $f(x)$ not intersects x - axis between $0$ to $1.$ Thus the second time $f(x)$ intersects x axis is before $0.$

Thus only one real and positive solution exists.

we can draw the graph of x^4 and 4x+1, for x to be positive , they have only one intersection , so the number of solution is 1

By Descartes rule maximum no. of positive roots = no. of sign changes =1

Also at x=0 ; y=-1

And for x>1 ; x^4 increases at a rate greater than that of 4x

therfore the no. of positive real root can not be 0 , therefore no. of positive real roots =1

as the max no. of positive roots in any polynomial= no. of sign change between coefficients then x^4-4x-1=0 has only one time sign changing so it has 1 positive real root

we can't say about the nature of the roots by this

This can also be solved by Descartes Rule (approximation of roots) , by counting the number of sign changes in the function

Here, f(x) = x^4 - 4x -1

Number of sign changes = 1 , therefore maximum number of positive real roots is 1

To find out maximum number of negative real roots, observe the sign changes in f(-x) infact that also comes out to be 1

The equation is changing the sign only once. So, only one positive solution. As simple as it is.

since there is only one sign change in $f(x)$ i.e, $+--$ so there will be only one positive real root

Using Descartes' Rule of Sign is the easiest way to solve this question.