# $x^2 + (k-3)x + k = 0$, ranges of k for roots to be of same sign

I need some help on the following.

The quadratic that I am dealing with is $x^2 + (k-3)x + k = 0$, and I need to find ranges of values of $k$, for which the roots will have the same sign.

For the roots of $ax^2+bx+c=0$ to have same signs, $a(x^2+ \frac{b}{a}x+\frac{c}{a})$, the last term, i.e. $\frac{c}{a} > 0$, because if you factorize the quadratic, to arrive at positive constant you either have to have two negative numbers multiplied or two positive multiplied by each other.

I will show what I have done.

I first thought that the roots will have same sign if they are the same, i.e. $b^2-4ac=0$, that led me to the result that $k=1$ or $k=9$.

Then I looked at $k$, $k>0$, according to what I said.

And the answer is actually $0<k<1$ and $k>9$. I have noticed that for this quadratic to have real and distinct roots, it has to satisfy following $k<1$ or $k>9$. Please help me to arrive at the required result. Thank you

• Oh I suppose I just combine my two answers... I.e. The roots have to be real and distinct and k>0. And that gives the answer. But what about the cases when k=1 and k=9, those seem to be valid too. In which case, the book answer should have been $0 < k \leq 1$ and $k \geq 9$ hmmm – i squared - Keep it Real Jul 21 '15 at 19:23
• @i squared - Keep it Real. Maybe the book requires the roots to be different. – wythagoras Jul 21 '15 at 19:45
• @wythagoras nope. I noticed it consistently omits the equal sign. This is "the core course for a level mathematics" by bostock and chandler, exercise 3d, 3. – i squared - Keep it Real Jul 21 '15 at 20:32

We solve the equation: \begin{align} x^2+(k-3)x+k = 0 &\implies x = \frac{3-k \pm \sqrt{k^2-6k+9 - 4\cdot 1\cdot k}}{2} \\ &\implies x = \frac{3-k \pm \sqrt{k^2-10k + 9}}{2}\end{align} For starters, we must have $k^2-10k + 9 = (k-1)(k-9) \geq 0$, which happens if $k \leq 1$ or $k \geq 9$. Recall that two numbers have the same sign if and only if their product is positive, and by Vieta's formulas, their product is $k$. So far, we have $0 < k \leq 1$ or $k \geq 9$. However, $k=1$ gives $1$ as a double root and $9$ gives $-3$ as a double root. So the book forgot these cases: our final answer is $0 < k \leq 1$ or $k \geq 9$.
There are two roots provided that the discriminant is positive, hence when: $$(k-3)^2-4k = k^2-10k+9 = (k-1)(k-9) > 0$$ so if $k<1$ or $k>9$. By Vieta's theorem the midpoint of the roots is $\frac{3-k}{2}$, hence if both the roots are positive $k$ must be less than $3$. $k<1$ is a stronger condition: to have two positive roots, now we just need to require that the original polynomial evaluated in zero is positive. So we have two positive roots when: $$0<k<1$$ and two negative roots when: $$k>9.$$