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We know it is easy to decide whether a quadratic equation has a root in a field which has characteristic $\neq$ 2. It is equivalent to judge whether there exists an element whose square is a specific number (cancel the linear term).

But this method doesn't work for the case of characteristic 2. Is there any idea to solve this case?

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    Why does it not work? A quadratic polynomial, say, $x^2+ax+b$ has no root over $K$ if and only if it irreducible. Also in characteristic $2$. – 2017-01-22
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    @Dietrich, maybe the point of the question is that in other characteristics you just have to check whether $a^2-4b$ is a square, but this breaks down in characteristic 2. – 2017-01-22
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    @GerryMyerson Yes, you got it. – 2017-01-22

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The quadratic equation $x^2+ax+b$ in the case of characteristic 2 takes two forms depending on whether or not $a=0$. When $a=0$, it's clear if $b$ is not a square there is no solution over the ground field, and the quadratic factors as $(x+\sqrt{b})^2$ over an extension of the ground field generated by attaching $\sqrt{b}$.

When $a\neq0$ we can show that the irreducible quadratic $x^2+ax+b$ cannot have a solution $\alpha = u+v\sqrt{w}$ with $u,v,$ and $w$ in the ground field. Suppose it did, then $$ \begin{align} 0 &= (u+v\sqrt{w})^2 + a (u+v\sqrt{w}) + b \\ 0 &= u^2+wv^2+au + v\sqrt{w}+b \\ v\sqrt{w} &= u^2+wv^2+au+b \end{align} $$ which is impossible, since $\sqrt{w}$ cannot be in the ground field.

While a solution using radical extensions is impossible, the case $a\neq 0$ can be reduced to considering equations of the form $x^2+x+c$. David Cox in his book Galois Theory calls a root of this equation a $2$-root of $c$ and denotes it $R(c)$. It's easy to check that $R(c)+1$ also satisfies the equation, so we now have two $2$-roots of $c$. The "quadratic formula" for characteristic 2 becomes $$ x= aR(b/a^2), a(R(b/a^2)+1) \text { provided }\, a\neq 0 $$