When does the equations have $1,2,3$ solutions?

Question

There is given an equation,

$$ \frac{x^2-x+1}{x^2+x+1}=kx+1$$

When does this solution have one, two, three real solution(s) in $x$.

My Approach:

The above equation can be rewritten as,

$$\frac{-2x}{x^2+x+1}=kx$$

This forms a cubic equation in $x$ where $0$ is a universal solution. So, then canceling $x$ gives a quadritic equation and which can add either $0,1,2$ more solutions according to value of discriminant of equation ($D<0,D=0,D>0$) respectively.

But this gave,

• $k\in \bigg(-\infty,\frac{-8}{3}\bigg)\cup[0,\infty)$ for one solution

• $k=\frac{-8}{3}$ for two solutions

• $k\in\bigg(\frac{-8}{3},0\bigg)$ for three solutions

But my textbook has different answer for second and third part. Where Am I wrong? And if you know different method to this problem please post that, I am glad to know alternatives.

Answer 1

Another approach to this problem is the graphical method. It isn't hard to plot a rough graph for $f(x)=\frac{x^2-x+1}{x^2+x+1}$. Which is shown below,

It touches $y=1$ at $x\rightarrow±\infty$.

Note that $g(x)=kx+1$ has the same $y-$intersept as that of $f(x)$.

Let for $k=k_1$ $g(x)$ is tangent to $f(x)$.

Appling condition of tangency gives,

• $k_1=-2$ at $x=0$ and cuts again at $x=-1$ like this,

• $k_1=-\frac{8}{3}$ at $x=-\frac{1}{2}$ and obviously cuts at $x=0$ like this,

So, looking at the graph for intersects will give,

1. $k\in\bigg(-\infty,-\frac{8}{3}\bigg)\cup[0,\infty)$, for one solution

2. $k=-\frac{8}{3},-2$, for two solutions

3. $k\in\bigg(-\frac{8}{3},0\bigg)-\{2\}$

Mistake in your solution:

at $k=-2$ there will be three roots $x=0,\{0,-1\}$ but two solutions which you counted three as per quadritic.

Answer 2

Your method is great. But you did not notice $x=0$ for the second and third case. For $kx^2+kx+k+2=0$, if $k=-2$, we would get a solution $x=0$, which you cancelled out at the first step. In this case, you would have two solutions.