1. range of values

Find the range of values of k if $\displaystyle x^2+(k-3)x+k=0$ has roots with the same sign .

My working :
$\displaystyle x^2+(k-3)x+k=0$
let a and b be two roots of this equation .
If both a and b were positive , (a+b)>0 , thus -k+3>0 , and 3>k
Similarly ab> 0 , thus k>0

If both a and b were negative , (a+b)<0 , thus -k+3<0 , then 3<k
Similarly , ab>0 , thus k>0

So the range of values should be 3>k , k>0 , and 3<k .
But the answer given is k>0 only . So where is my mistake ?

Find the range of values of k if $\displaystyle x^2+(k-3)x+k=0$ has roots with the same sign .

My working :
$\displaystyle x^2+(k-3)x+k=0$
let a and b be two roots of this equation .
If both a and b were positive , (a+b)>0 , thus -k+3>0 , and 3>k
Similarly ab> 0 , thus k>0

If both a and b were negative , (a+b)<0 , thus -k+3<0 , then 3<k
Similarly , ab>0 , thus k>0

So the range of values should be 3>k , k>0 , and 3<k .
But the answer given is k>0 only . So where is my mistake ?
Well suppose the result give is correct, then what happens when $\displaystyle k=3$ (the only point that your answer suggests the roots either don't exist or have different sign) ?

The quadratic reduces to $\displaystyle x^2+3=0$ which has no real roots, so the given answer is obviously wrong at the only point it disagrees with yours.

CB