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 $k<3$.
Best Answer
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. $k<3$.