(Discriminant) For which values of k will the equation g(x) = x + k have two real roots that are of opposite signs?

Question

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 $k<3$.

Answer 1

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. $k<3$.

Answer 2

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 $k<3$.

Answer 3

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

$k^2-4k+12>k^2$

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?

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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.