Let $K$ be a field with $\operatorname{char}K \neq 2$. Suppose that $p(x)=x^2-a$ is irreducible over $K$ a let $\lambda$ a root of $p$. Show that $x^2 - \lambda$ has root in $K(\lambda) \iff $ there exist $x \in K$ s.t $-4a = x^4.$
I'm studying galois theory and I have found this exercise.
Well, $x^2-a$ is irreducible over $K$, so $\{1, \lambda\}$ is a basis of $K(\lambda)$ over $K$.
$(\Rightarrow):$
Suppose that $\sqrt {\lambda} \in K(\lambda)$, so there exists $x,y \in K$ s.t $\sqrt{\lambda} = x +y \lambda$, also we can say $y \neq 0$, because $\sqrt{\lambda} \notin K$.
Then we have
$$\lambda = (x+y\lambda)^2 = x^2+ 2xy\lambda + y^2a$$.
Comparing the coefficients we have $x^2 + y^2a = 0$ and $2xy = 1$. Since $\operatorname{char}K \neq 2$, we have $x= (2y)^{-1}$, so $\frac{1}{4} + y^4a = 0$, that is, $-4a = (y^{-1})^4$.
$(\Leftarrow):$ How can I show the otherside???
For the other side, if $-4a = x^4$, then $\frac x 2 + \frac \lambda x$ is a square root of $\lambda$, because $$\left(\frac x 2 + \frac \lambda x\right)^2 = \frac{x^2} 4 + \frac{\lambda^2}{x^2} + \lambda = \frac{x^4 + 4a} {4x^2} + \lambda = \lambda.$$