I am currently in Grade 12 and came across the following question in a past paper:

$$g(x) = \frac{2}{x+1}+1$$

The question asks: For which values of k will the equation $g(x) = x + k$ have two real roots that are of opposite signs?

After simplifying the equation I come to : $x^2 + kx + (k-3) = 0$

From the question i know to use the discriminant ($b^2 -4ac$) and I know that the discriminant of the function must be greater than zero for two real solutions, however i am unsure as how to have the roots to be of opposite signs.

Nevertheless I continued to simplify the inequality and i came to: $k^2 - 4k + 12$ is greater than zero.

from this step i am unable to factorise and thus i am unable to solve the inequality.

I would appreciate any guidance as to how to get the roots to be of different signs and how to solve the ineqaulity.

The answer according to the memo is that $k<3$.


The discriminant of the equation $x^2+kx+(k-3)=0$ is $k^2-4k+12=(k-2)^2+8$, which is positive, so the roots of the equation are both real, for any choice of $k$.

Let the roots be $r_1$ and $r_2$. Writing $x^2+kx+(k-3)=(x-r_1)(x-r_2)$ and expanding, we see the product of the roots is $k-3$. We want this product to be negative, i.e. $k<3$.


So you know the discriminant is $k^{2}-4k+12$. If we complete the square, we see that $k^{2}-4k+12\equiv (k-2)^{2}+8$. As $k-2$ is a real number, it follows that $(k-2)^2\geq 0$, so that $(k-2)^{2}+8\geq 8>0$. Hence, the discriminant is always positive no matter what value $k$ takes, and we're done with this part of the question: we know that the equation will have two (distinct!) real roots.

Let's denote the roots of the equation $g(x)=x+k$ by $\alpha$ and $\beta$. Then, by your own working, we must have $(x-\alpha)(x-\beta) \equiv x^{2}+kx+(k-3)$ (notice that this is an identity, not just an equation, so it holds for all $x$). Then, we must have $x^2-(\alpha+\beta)x+\alpha\beta \equiv x^{2}+kx+(k-3)$. Comparing coefficients, we see that $\alpha\beta=k-3$.

Now, suppose the roots are of opposite signs: then we must have $\alpha\beta < 0$; but $\alpha\beta = k-3$, so we must have $k-3<0$, i.e., $k<3$. Going the other way, if we assume $k<3$, it follows that $\alpha\beta<0$, and so the roots must be of opposite signs.

Hence, the roots are of opposite signs if and only if $k<3$.


Note that to get roots to be of each sign would require:


The first part of the Quadratic Formula where the square root of the discriminant would have to be greater than the $b$ term in the formula which is $k$ in this case.

From the above inequality, it is easy to get $k<3$ as a result, right?

Consider 2 values, $x+y$ and $x-y$. Now, for one to be positive and one to be negative, $y>x$ must be true or else both will sure whatever sign $x$ has. If necessary, plug in numbers to see this point as it is rather basic algebra to my mind.

The leap from this to the inequality I have above is that $k$ and $\sqrt{k^-4k+12}$ are each squared first.

  • $\begingroup$ To be honest I don't understand, firstly why would the inequality need to be greater than k^2, please elaborate, if possible could you please suggest a video, book or website that explains this concept. $\endgroup$ – Mohamed Ameen Omar Jul 17 '15 at 0:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.