The quadratic equation $x^2+x=3kx-k^2$ has two different real roots. Find the range of $k$ 
The quadratic equation $x^2+x=3kx-k^2$ has two different real roots. Find the range of $k$.

My answer is $k<1$ or $k<\frac{1}{5}$, but the answer sheet says $k<\frac{1}{5}$ or $k>1$.
What have I done wrong? Please help.
What I've done
$(1-3k)^2-4(k^2)>0$
$1-6k+9k^2-4k^2>0$
$\frac{6 \pm 4}{10}>k$
$k<1$ or $k<\frac{1}{5}$

The discriminant of $$x^2+x(1-3k)+k^2=0$$
$$(1-3k)^2-4k^2=1-6k+5k^2=5\left(k-\dfrac15\right)(k-1)$$ which needs to be $\ge0$
Now for $(y-a)(y-b)\ge0$ with $a\le b$
If $y-a\ge0\iff y\ge a;$ we need $y-b\ge0\iff y\ge b\implies y\ge$ max$(a,b)=b$
What if $y-a\le0?$
