Finding $b$ such that $\dfrac{1}{\left(x - 2\right)^2} + b = -x$ has exactly two solutions, using a non-calculus approach

Question

The question:

Let $f(x) = \dfrac{1}{\left(x - 2\right)^2} + b.$

Find $b$ when $f(x) = -x$ has only two solutions.

My attempt at a solution (without using any calculus):

$$\dfrac{1}{\left(x - 2\right)^2} + b = -x$$ $$\dfrac{1}{\left(x - 2\right)^2} = -x - b$$ $$1 = \left(x - 2\right)^2\left(-x - b\right)$$ $$1 = -x^3 + bx^2 + 4x - 4bx + 4b - 4b$$

I end up with a cubic polynomial which complicates finding any discriminant.

I used calculus by setting $\dfrac{d}{dx}\left(-x\right) = \dfrac{d}{dx}\left(f(x)\right)$, and found $b = -\left(\dfrac{3}{2^{\frac{2}{3}}} + 2\right)$, however I want an purely non-calculus approach.

Any help would be greatly appreciated!

Edit:

I translated the graphs in the negative x direction by two units, which gave me: $$f(x) = \dfrac{1}{x^2} + b$$

$$\textrm{and therefore}, f(x) = -x - 2$$

This gave me the cubic equation $x^{3}+\left(2+b\right)x^{2}+1=0$, however I am still unsure to find a value of $b$ algebraically that would give me two solutions.

Answer 1

We want the polynomial $x^3 + (2 + b)x^2 + 1$ to have exactly two different, non-zero (in the beginning we worked with expression that involved division by $x^2$), real roots. This can be denoted as: $$x^3 + (2 + b)x^2 + 1 = (x-A)(x-B)\cdot Q(x)$$where $A \neq B$ and $Q(x)$ has only roots that belong to the set $\{A, B, 0\}$ (even if $0$ was the root of $Q(x)$ the original equation would still have only two solutions, because $0$ is not in the domain of original equation).

Moreover we can see that $Q(x)$ must be polynomial of degree $1$. $Q(x)$ cannot have root equal to $0$, since $0$ is not the root of $x^3 + (2 + b)x^2 + 1$. So we can denote (without loss of generality) $Q(x) = (x - B)$.

Now by expanding the expression we get: $$ (x-A)(x-b)^2 = x^3 + (-2B -A)x^2 + (B^2 + 2AB)x - AB^2 $$ We can compare the coefficients of that polynomial with the coefficients of the original one: $$ \begin{cases} -2B - A = 2 + b \\ B^2 + 2AB = 0 \\ -AB^2 = 1 \end{cases} $$ If we divide second equation by $B$ we will get: $$ B + 2A = 0 \iff B = -2A $$ Plugging it to third one will result in: $$ -4A^3 = 1 \iff A = \frac{-1}{\sqrt[3]{4}} $$ Now: $$ B = \frac{2}{\sqrt[3]{4}} $$ And by plugging it to first equation will give us the value of $b$: $$ b = \frac{-4 + 1}{\sqrt[3]{4}} - 2 = \boxed{\frac{-3\sqrt[3]{2}}{2} - 2} $$

Answer 2

Let $a:=b+2$. Since $0$ is not a root of $x^3+ax^2+1=0$, the latter has exactly two solutions iff $1+ay+y^3=0$ does (substituting $y=\frac1x$), i.e. iff there exist two distinct numbers $\alpha,\beta$ such that$$Y^3+aY^2+1=(Y-\alpha)^2(Y-\beta). $$This is equivalent to $$\begin{cases}2\alpha+\beta&=0\\\alpha^2+2\alpha\beta&=a\\\alpha^2\beta&=-1\end{cases}$$ i.e. $$\begin{cases}\beta&=-2\alpha\\a&=-3\alpha^2\\\alpha^3&=\frac12.\end{cases}$$ The answer is therefore $$b=-\frac3{\sqrt[3]4}-2.$$

Answer 3

The answer to this problem can be written in terms of $x$, solution of the following equation: $$x\bigg( 1 + \frac{x}{r} \bigg)^r = x\cdot e_r(x)= z$$ assuming $r \in Q$ and $z \in C$. The solution $x$ is the so-called Lambert-Tsallis function, $W_r(z)$, proposed by R. V. Ramos (please see some answers in my profile). Also, $[e_r(x)]^\alpha = e_{r\alpha}(x\alpha)$. As can be seen, $W_r(z)$ is a multivalued function and it has a lot of interesting properties.

First, let's call $\color{blue}{ x+b=y}$, then $$\bigg[ y - (2+b)^{-2}\bigg] = -y$$ $$= y^{-1}\bigg\{ -(2+b)\big[ 1 - \frac{y}{2+b}\big] \bigg\}^{-2} = -1$$ $$= y^{-1}\frac{1}{(2+b)^2}\bigg[ 1 - \frac{y}{2+b} \bigg]^{-2} = -1$$ $$= y^{-1}e_{-2}\bigg( \frac{2y}{2+b} \bigg) = -(2+b)^2$$ raising to $\color{red}{r=\frac{1}{-1}}$ we have $$= y^{-r}e_{-2r}\bigg( \frac{2r}{2+b}y \bigg) = \big[-(2+b)^2\big]^r$$

then we find:

$$y^{-r}=\frac{W_{-2r}\bigg\{ \frac{2r}{2+b}[-(2+b)^2]^{r}\bigg\}}{\frac{2r}{2+b}} \text{ }$$ Using the properties of Lambert-Tsallis function and Tsallis exponential, let's raise again to $^1/r$ to find

$$y^{-1}=-(2+b)^2 \frac{1}{e_{-2}\bigg[\frac{W_{-2r}(z)}{r}\bigg]} $$ with $z= \frac{2r}{2+b}[-(2+b)^2]^{r}$. $$=y^{-1}=-(2+b)^2 \frac{1}{\bigg[1-\frac{W_{-2r}(z)}{2r}\bigg]^{-2}} $$ $$=\frac{1}{x+b}= \frac{-(2+b)^2}{\bigg[1-\frac{W_{-2r}(z)}{2r}\bigg]^{-2}} $$

resulting on the final closed expression to $x$

$$\color{blue}{x= -b - \frac{(2+b)^{-2}}{\bigg[1-\frac{W_{-2r}(z)}{2r}\bigg]^{2}}} .$$

Now, let's go to the funny part... $r=-1$, so

$$\color{blue}{x= -b - \frac{(2+b)^{-2}}{\bigg[1+\frac{W_{2}(z)}{2}\bigg]^{2}} \text{ with } z= +2(2+b)^{-3}}.$$

As can be seen $x$ can be written in terms of $W_2(z)$. The Lambert-Tsallis function with $r=2$ provides 3 values, as expected if we look at the original problem.

However, when the subindex of Lambert-Tsallis belongs to natural set there is a property which provides that in $z=- \bigg(\frac{r}{r+1}\bigg)^{r+1}$ there is a double root ($-r/(r+1)$). The subindex is $-2r=2.$

Therefore, in our final expression to $x$, there will be a double root in $x$ when $z= +2(2+b)^{-3} = -\bigg(\frac{2}{3}\bigg)^3$ resulting on the final answer $\color{blue}{b= -2 -\frac{3}{4^{1/3}}}$

Although this solution could be new to everyone, it provides you the analytical expression to $x$ if you are interested.... So let's continue

The analytical values of $W_2(z)$ at the double root is $-2/3$ (the double root) and $-8/3$ (calculated using r=2- subindex - by $-r^2 -2 \cdot \frac{-r}{r+1}$) so

$$\color{blue}{x= -b - (2+b)^{-2} \cdot {[ 9/4 ; 9/1]}}.$$ resulting at

x={3.2599 ; 1.37}