For what value of k, $x^{2} + 2(k-1)x + k+5$ has at least one positive root?

Question

For what value of k, $x^{2} + 2(k-1)x + k+5$ has at least one positive root?

Approach: Case I : Only $1$ positive root, this implies $0$ lies between the roots, so $$f(0)<0$$ and $$D > 0$$

Case II: Both roots positive. It implies $0$ lies behind both the roots. So, $$f(0)>0$$ $$D≥0$$ Also, abscissa of vertex $> 0 $

I did the calculation and found the intersection but its not correct. Please help. Thanks.

Answer 1

You only care about the larger of the two roots - the sign of the smaller root is irrelevant. So apply the quadratic formula to get the larger root only, which is $\frac{-2(k-1)+\sqrt{4(k-1)^2-4(k+5)}}{2} = -k+1+\sqrt{k^2-3k-4}$. You need the part inside the square root to be $\geq 0$, so $k$ must be $\geq 4$ or $\leq -1$. Now, if $k\geq 4$, then to have $-k+1+\sqrt{k^2-3k-4}>0$, you require $k^2-2k-4> (k-1)^2$, which is a contradiction. Alternately, if $k\leq -1$, then $-k+1+\sqrt{k^2-3k-4}$ must be positive, as required.

So you get the required result whenever $k\leq -1$.

Answer 2

The roots are given by $1-k\pm\sqrt{k^2-3k-4}$, for which we have: $$\cases{ 1 - k + \sqrt{k^2 - 3k - 4} > 0 & if $\phantom{~-5< \;}k \le -1$\\ 1 - k - \sqrt{k^2 - 3k - 4} > 0 & if $~-5<k\le -1 $. } $$ Wolfram Alpha gives the plus and subtract cases. So for $k\le -1$, you get at least one positive root.

Answer 3

Suppose $x_{1}$ is a real root, then we have that:

$$ (x_{1}+(k-1))^{2} - (k^2-3k-4) = 0 $$ $$(x_{1}+(k-1))^{2} = (k^2-3k-4)$$ $$(k^2-3k-4) \ge 0$$

It's obviously seen that the positive roots are got only when $k \le -1$.

Q.E.D.