If the graph of the function $f(x)=x^2+2(k-1)x+k+5, k\in R$ cut the x-axis at least at one point on the positive side , find the set of possible values of the constant k.
My attempt is as follows:
As graph is said to be cutting the x-axis, it means $D>0$
$$4(k-1)^2-4*(k+5)>0$$ $$k^2-3k-4>0$$ $$(k-4)(k+1)>0$$ $$k\in (-\infty,-1) \cup (4,\infty)$$
Case 1: If one root is negative and other is positive.
It means $f(0)<0$
$$k+5<0$$ $$k<-5$$ $$k\in (-\infty,-5)$$
Case 2: If one root is greater than equal to zero and other root is positive.
So $f(0)>=0$ and $0<\frac{a+b}{2}$ where a and b are roots.
$$k+5>=0 \quad \cap \quad 0<\frac{-2(k-1)}{2}$$ $$k>=-5 \quad \cap \quad k<1$$ $$k\in [-5,-1)$$ (took intersection with discriminant also)
Taking union of case 1 and case 2, gives us $k\in (-\infty,-1)$, but answer is $k\in (-\infty,-1]$
Where am I making the mistake? I took $D>0$ as in the question it is said that graph cuts the x-axis, so it means that are definitely two roots of the equation.
The "mistake" (actually, really a difference in interpretation of a term) is the question asks for it to "cut" the $x$-axis. Note that as saulspatz's comment states, the concept of "cut" is somewhat vague, and not a technically defined term. Nonetheless, just touching the $x$-axis is, in effect, cutting it into $2$ parts, the part before where the graph touches it and the part after it. This seems to be what the questioner is intending.
As such, this means there can be either one or two roots, so a repeated root, as long as it's positive, is allowed. Thus, the test is not for $D \gt 0$ but for $D \ge 0$. Using this, you'll get an end result which includes $-1$ to end up with $k \in (-\infty, -1]$, as the answer states.