Disposition of roots

Graphs are really amazing

• The LHS of function can be broken into two separate parts $\frac{4x}{x^{2}+x+3}=p$ and $\frac{5x}{x^{2}-5x+3}=q$ where $p+q=-\dfrac{3}{2}$

Note : One can easily plot $p$ and $q$ separately (By checking sign of derivative and all..)

• Now we see that for $x \geq 0$ the expression $p+q$ can get $-2.359$ (adding extremes of both graphs for $x \geq 0$ ) as the nearest value to $-1.5$ means no solution for $x \geq 0$

• For $x \leq 0$ we see that $p+q$ can get the $-2.214$ as the least value (adding minima of both graphs for $x \leq 0$ ) and will increase on its left and right upto $0$ .And will look like:

So there are $\color{#D61F06}{2}$ real solns. and $\color{#D61F06}{2}$ imaginary.

Now to check for rational roots lets apply Rational Root Theorem the complete equation can be transformed as:

${f(x)= \, \, \dfrac{3 \; x^{4} + 6 \; x^{3} - 27 \; x^{2} + 18 \; x + 27}{2 \; x^{4} - 8 \; x^{3} + 2 \; x^{2} - 24 \; x + 18}} =0$

Now $3x^{4} + 6 \; x^{3} - 27 \; x^{2} + 18 \; x + 27 =0$ the possible rational roots can be $-9,-3,-1,1,3,9$ by substituting all we can see that none of them is a solution.

One should feel the essence of graphs they are really the most powerful tool in Mathematics.Just use them as much as you can!!!

---

Is there any other method