Does there exist a field $K$ and $x \in K$ with $K(x^{1/4})=K(x^{1/2}) \neq K$ I want to know if there exists a field $K$ and $x \in K$ with $K(x^{1/4})=K(x^{1/2}) \neq K$.

My attempt thus far: if $K(x^{1/2}) \neq K$ then $x^{1/2} \notin K$ so the minimal polynomial of $x^{1/2}$ over $K$ has degree $&gt;1$ and is therefore $X^2-x$. Then $[K(x^{1/2}):K] = 2$ and $\{1, x^{1/2}\}$ is a $K$-basis for $K(x^{1/2})$. Since $K(x^{1/4})=K(x^{1/2})$ we have $x^{1/4} \in K(x^{1/2})$ so there exist $a, b \in K$ such that $x^{1/4} = a + bx^{1/2}$.

Then I tried rearranging and squaring both sides etc to see if I could find something useful but I didn't get anywhere... Any tips? I don't even know whether such a $K$ exists at all!
Yes. $K= \mathbb R$, $x=-1$.

Added: Generalising comments by k.stm and Jyrki Lahtonen, as well as reuns under the OP, I claim that assuming $char(K) \neq 2$, such an $x \in K$ exists if and only if $-1$ is not a square in $K$. The proof, as well as the case $char(K)=2$, are left as exercise.
$s=x^{1/4}$ , $d= x^{1/2}$

s£ k(d)==> s=a+bd ===> s^2 =a^2 +b^2d^2 +2abd ==> d=a^2+b^2x+2adb 
===> d( 1-2a) = a^2 +x 
d= ( 1-2ab)^(-)  ( a^2+xb^2)  , if 1-2ab#0 ==> d£K but this absurde since  k( d) #k  sorry i have note Latex for better redaction