number of real roots of complex cubic equation

Question

If $a,b,c$ are distinct real numbers.

Then number of real roots of the equation

$$\frac{1}{x-a}+\frac{1}{x-b}+\frac{1}{x-c} = \frac{1}{(x-a)(x-b)}+\frac{1}{(x-b)(x-c)}+\frac{1}{(x-c)(x-a)}$$

Plan

$$(x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=(x-a)(x-b)(x-c)$$

$$x^3-(a+b+c)x^2+\sum ab x-abc=3x^2-2(a+b+c)x+ab+bc+ca$$

$$x^3-(a+b+c+3)x^2+(\sum ab+\sum a)x-\sum ab=0$$

How do i solve it help me

Answer 1

It should be

$(x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=(x-a)+(x-b)+(x-c)$

which is quadratic.

This can be written as

\begin{align*} \left[(x-b)(x-c)-\frac12(x-b)-\frac12(x-c)+\frac14\right]\qquad\quad&\\ +\left[(x-c)(x-a)-\frac12(x-c)-\frac12(x-a)+\frac14\right]\quad&\\ +\left[(x-a)(x-b)-\frac12(x-a)-\frac12(x-b)+\frac14\right]&=\frac34\\ \left(x-b-\frac12\right)\left(x-c-\frac12\right)+\left(x-c-\frac12\right)\left(x-a-\frac12\right)\qquad\quad&\\+\left(x-a-\frac12\right)\left(x-b-\frac12\right)&=\frac34\\ \end{align*}

Let $y=x-c-\frac12$, $c-a=p$ and $c-b=q$. The equation can be then written as

\begin{align*} (y+c-b)(y)+(y)(y+c-a)+(y+c-a)(y+c-b)&=\frac34\\ 3y^2+2(p+q)y+pq-\frac34&=0 \end{align*}

The discriminant is

$4(p+q)^2-4(3)(pq-\frac34)=4p^2-4pq+4q^2+9=(2p-q)^2+3q^2+9>0$

The equation has two real roots.

Answer 2

You may want to check your derived equation. When I multiply the given equation by $(x-a)(x-b)(x-c)$ I get

$(x-b)(x-c)+(x-a)(x-c)+(x-a)(x-b)=(x-c)+(x-a)+(x-b)$

$3x^2-(2a+2b+2c+3)x+(bc+ac+ab+c+a+b)=0$

You should be able to show that the discriminant is

$4(a^2+b^2+c^2)+4(ab+ac+bc) +9$

=$(2a^2+4ab+2b^2)+(2a^2+4ac+2c^2)+\text{two more nonnegative terms}>0$

and infer the number of roots of the quadratic accordingly.

Answer 3

As has been said by CY Aries and others, multiplying with $(x-a)(x-b)(x-c)$ yields a quadratic equation in $x$, not a cubic one:

$$(x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=(x-a)+(x-b)+(x-c) \tag{1}\label{eq1}$$

However, $\eqref {eq1}$ and your original equation are not equivalent!

Your original equation has terms that are undefined for $x=a, x=b$ and $x=c$, so a solution $x$ must be found in the set $\mathbb R\backslash\{a,b,c\}$. Equation $\eqref {eq1}$ however has no such undefined terms, so the solutions are usually sought in the full $\mathbb R$. Making a statement about how many solutions (in $\mathbb R$) $\eqref{eq1}$ has is not a complete solution for your problem, if it isn't checked that any such solutions are different from $a,b$ and $c$.

Continuing the same way as Oscar Lanzi, we expand the terms and then collect the factors before the different powers of $x$:

$$3x^2-(2a+2b+2c+3)x+(ab+ac+bc+a+b+c)=0. \tag{2}\label{eq2}$$

Note that $\eqref{eq1}$ and $\eqref{eq2}$ are actually equivalent over $\mathbb R$, so solving $\eqref{eq2}$ is exactly the same as solving $\eqref{eq1}$.

The number of solutions in the real numbers for such a quadratic equation is dermined by its discriminant $$\Delta = (2a+2b+2c+3)^2-4\times3(ab+ac+bc+a+b+c). \tag{3}\label{eq3}$$

If $\Delta <0$, $\eqref{eq2}$ has no real solution, if $\Delta = 0$ it has exactly one (double) real solution and if $\Delta > 0$ it has exactly 2 different real solutions.

Simplifying $\eqref{eq3}$ yields

$$\begin{eqnarray} \Delta &=& (4a^2+4b^2+4c^2+9+8ab+8ac+8bc+12a+12b+12c)-12(ab+ac+bc+a+b+c) \\ &=& 4a^2+4b^2+4c^2-4ab-4ac-4bc+9\\ &=& (2a^2-4ab+2b^2) + (2a^2-4ac+2c^2) + (2b^2-4bc+2b^2) + 9\\ &=& 2(a-b)^2+2(a-c)^2+2(b-c)^2+9\\ &>& 0. \end{eqnarray} $$ This means $\eqref{eq2}$ (and therefore $\eqref{eq1}$) always has exactly 2 different real solutions.

But as I said in the beginning, we are not done yet! It could be that one or both solutions are equal to $a,b$ or $c$.

Checking that is easier in $\eqref{eq1}$, when we substitute $x=a$ we get after remving the $(a-a)$ terms:

$$(a-b)(a-c)=(a-b)+(a-c)$$

That is obviously not always true, but is also certainly not impossible to happen, for example it's true for $$a=1.5,b=0,c=-1.5$$

Of course, the check must be made to exclude $b$ and $c$ from the solutions of $\eqref{eq1}$ as well, so the full set of conditions to check is

$$ \begin{eqnarray} (a-b)(a-c) &=& (a-b)+(a-c)\\ (b-a)(b-c) &=& (b-a)+(b-c) \tag{4}\label{eqcond}\\ (c-a)(c-b) &=& (c-a)+(c-b)\\ \end{eqnarray} $$

Obviously, almost all sets of parameters $(a,b,c)$ will not fullfill any of the 3 conditions in $\eqref{eqcond}$, so then $\eqref{eq1}$ has exactly 2 different real solutions. If exactly one of the 3 conditions in $\eqref{eqcond}$ is fullfilled, as for $a=1.5,b=0,c=-1.5$, then $\eqref{eq1}$ has exactly 1 real solution. It needs to be checked if there could even be parameter sets $(a,b,c)$ where 2 conditions are met, that means where both solutions to $\eqref{eq2}$ are in the set $\{a,b,c\}$.

Note that the conditions in $\eqref{eqcond}$ are symmetric in $\{a,b,c\}$ (because $\eqref{eq1}$ is symmetric in $\{a,b,c\}$), so it is enough the check if the first 2 conditions can be fullfilled at the same time. Also note that $\eqref{eqcond}$ contains the variables only in pairwise differences, so it makes sense to define

$$r:=a-b, s:=b-c,$$

which transforms the first 2 conditions of $\eqref{eqcond}$ into

$$ \begin{eqnarray} r(r+s) &=& r+(r+s) \tag{5}\label{eq5}\\ -rs &=& -r+s \tag{6}\label{eq6}\\ \end{eqnarray} $$

Adding $\eqref{eq5}$ and $\eqref{eq6}$ gives $r^2=2s+r$ and thus

$$s=\frac{r^2-r}2. \tag{7}\label{eq7}$$

Putting this into $\eqref{eq6}$ yields $-r\frac{r^2-r}2=-r+\frac{r^2-r}2$ which, when multiplied by $-2$, is equivalent to $r^3-r^2=2r-(r^2-r)$ and thus

$$r^3-3r=0$$

Now this has 3 solutions $r_1=\sqrt{3}, r_2=-\sqrt{3}$ and $r_3=0$. $r_3$ is not suitable for us, as it would mean $a-b=0$ which is forbidden in the problem. From $\eqref{eq7}$ we get the $s$ solutions corresponding to $r_1,r_2$

$$s_1=\frac{3-\sqrt{3}}2,\,s_2=\frac{3+\sqrt{3}}2.$$

It's easy to check that those $(r_1,s_1)$ and $(r_2,s_2)$ are actually solutions of $\eqref{eq5}$ and $\eqref{eq6}$. Since $r_1+s_1=s_2$ and $r_2+s_2=s_1$, we also know that for both solutions $a-c=r+s \neq 0$.

Note that $c$ came out to be the smallest of the 3 parameters in both solutions, and both solutions differ just on which of the other 2 parameters you call $a$ and which one you call $b$. That means this all boils down to the following:

If the 3 parameters are orderd $a > b > c$ (which is w.l.o.g. possible as $\eqref{eq5}$ is symmteric with respect to any permutation of $\{a,b,c\}$) and we have $$ b-c = \frac{3-\sqrt3}2, \, a-c = \frac{3+\sqrt3}2$$ then $\eqref{eq1}$ has no real solution. Otherwise, if any 1 of the 3 conditions in $\eqref{eqcond}$ is fullfilled, $\eqref{eq1}$ has exactly one real solution. Finally, if none of the 3 conditions in $\eqref{eqcond}$ is fullfilled, $\eqref{eq1}$ has exactly 2 real solutions.

As confirmation, look at the following plot from Wolfram Alpha for the parameter set

$$a=\sqrt3, b=0, c=-\frac{3-\sqrt3}2,$$

which fullfills the 'no solutions' condition above:

The plot has a pole at $c$, but neither at $x=a$ and $x=b$ are poles, which corresponds with the fact that $\eqref{eq2}$ has 2 solutions at $x=a$ and $x=b$.