<s>[INST] 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 $k<\frac{1}{5}$
 [/INST] 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.
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