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Find the number of irrational roots of the equation $$\dfrac{4x}{x^2 + x + 3} + \dfrac{5x}{x^2 - 5x + 3} = -\dfrac{3}{2}.$$

I got a solution : divide both denominator and numerator by $x$. Let $x+\dfrac{3}{x} = y$. Then the equation becomes $$\dfrac{4}{y + 1} + \dfrac{5}{y - 5} = \dfrac{3}{2}.$$ Now simplifying this we get $y = -5, 3$.

Finally, $x+\dfrac{3}{x} = -5$ has 2 irrational roots, and $x+\dfrac{3}{x} = 3$ has 2 imaginary roots.

But my question is, since this question is for a competitive exam, is there any other quick approach to solve this question?

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    $\begingroup$ You found the quick approach (+1). $\endgroup$
    – dxiv
    Dec 31, 2021 at 7:30

1 Answer 1

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Here is another way to solve this problem. It is up to you and @dxiv to decide whether this method is longer or shorter than yours.

We start by shifting the $2^{\text{nd}}$ term on LHS of the given identity to its RHS. $$\dfrac{4x}{x^2 + x + 3} = -\dfrac{3}{2}-\dfrac{5x}{x^2 - 5x + 3}$$

When RHS is simplified, we have, $$\dfrac{4x}{x^2 + x + 3} = -\dfrac{3x^2 - 5x + 9}{2x^2 - 10x + 6}.$$

Now, we add the denominators to the respective numerators to get, $$\dfrac{x^2 + 5x + 3}{x^2 + x + 3} = \dfrac{-x^2 - 5x - 3}{2x^2 - 10x + 6}.$$

This implies, $$ x^2 + 5x + 3 = 0\qquad\text{and}\qquad x^2 + x + 3= -2x^2 + 10x - 6\quad\Longrightarrow\quad x^2 -3 x + 3=0.$$

The first quadratic equation gives us two irrational roots, while the second two imaginary roots

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