Help with $\sqrt{I}$, where $I=(y^2,x+yz)$ in $\mathbb{C}[x,y,z]$ 
$a)$ $\sqrt{I}$  where $I=(y^2,x+yz)$ in $\mathbb{C}[x,y,z]$.

first  it's clear  $y \in \sqrt{I}$ then $x=(x+yz)-yz \in  \sqrt{I}$ because  $yz \in \sqrt{I}$ 
is  it $\sqrt{I}=(x,y)$ ?

$b)$ $\sqrt{I}$  where $I=(x^2,xy^2)$ in $\mathbb{C}[x,y]$.

it's clear $x \in \sqrt{I}$ , $(xy)^2=x(xy^2)\in \sqrt{I}$ then $xy \in \sqrt{I}$
is  it $\sqrt{I}=(x,xy)$ ?
 A: You need to show the other direction. This follows from the fact that $\sqrt{I}$ is tne intersection of all prime ideals containing $I$. Now in the first case, $I \subseteq (x,y)$ we just need to show that $(x,y)$ is prime. This is clear since if $f$ and $g$ are such that $fg \in (x,y)$ then write $f=a(x,y,z)x+b(y,z)y +c(z)$ and 
$g=d(x,y,z)x+e(y,z)y +h(z)$ is is the clear that either $c(z)=0$ or $h(z)=0$.
For the second case, $(x,xy)=(x)$ and its clear this is prime.
A: An alternative viewpoint, from algebraic geometry:
$\mathbb{C}$ is algebraically closed, so by Hilbert's Nullstellensatz, we have, for any ideal $J$:
$\sqrt{J}=I(Z(J))$
Applying this to $a)$, clearly $Z(I) = \{(x,y,z): y = 0 = x\}$ and then it is clear again that $I(Z(I))=(X,Y)$.
Applying this to $b)$, $Z(I) = \{(x,y,z):x = 0\}$ so we immediately get $I(Z(I))=(X)$.
This approach may not seem quicker than the other methods presented here, and I suppose if you were to write out explicitly what I claim is "clear" then it probably wouldn't be. But even if you chose to do that, this method certainly isn't slower, and by thinking about problems this way it is often much easier to see what the answers. It also has the added advantage that you can avoid having to fiddle about with polynonials explicitly.
