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 $k<\frac{1}{5}$
 A: 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 $y-a\le0?$
A: We have that the discriminant $$\Delta _{f (x)} =(1-3k)^2-4k^2 =(5k-1)(k-1) $$ We can analyse this into three cases:
Case $(1) $: If $k <\frac {1}{5} $, then surely $(5k-1) $ and $(k-1) $ are negative. So, $\Delta $ is positive. (Acceptable)
Case $(2) $: If $k\in (\frac {1}{5},1) $, then $(5k-1) $ is positive but $(k-1) $ is negative. (Not acceptable)
Case $(3) $: If $k>1$, then both $(5k-1) $ and $(k-1) $ are positive. (Acceptable)
Using these cases, the result follows. Hope it helps.
A: The discriminant of this quadratic equation is $\Delta(k)=5k^2-6k+1$ and it has to be positive for the quadratic equation to have two real roots.
$\Delta(k)$ is a quadratic polynomial in $k$ and it has one obvious root: $k=1$, hence the other root is $\frac15$.
Now a well-known theorem asserts that a quadratic polynomial has the sign of its leading coefficient, except between its (real) roots, if any. So we can conclude that 
$$\Delta(k)>0\iff k<\frac15\enspace\text{or}\enspace k>1.$$
