Finding Four Roots

$x^4 + \dfrac{8}{9}x^2 + 1=3x^3 + 3x$

$\Rightarrow 9x^4 + 8x^2 + 9 = 27x^3 + 27x$

$\Rightarrow 9x^4 -27x^3 + 8x^2 - 27x + 9=0$

$\Rightarrow x^2\bigg[9x^2 -27x + 8 - \dfrac{27}{x} + \dfrac{9}{x^2}\bigg]=0$

$\Rightarrow x^2\bigg[9\bigg(x^2 + 2 +\dfrac{1}{x^2}\bigg) -27\bigg(x + \dfrac{1}{x}\bigg) + 8 -18\bigg]=0$

Since , $x \neq 0$ , $9\bigg(x + \dfrac{1}{x}\bigg)^2 -27\bigg(x + \dfrac{1}{x}\bigg) -10=0$

Let $x + \dfrac{1}{x} = t \Rightarrow 9t^2 - 27t - 10= 0 \Rightarrow (3t+1)(3t-10) =0$

$\Rightarrow t = \dfrac{-1}{3} or \dfrac{10}{3}$

When $t = \dfrac{-1}{3} , x + \dfrac{1}{x} =\dfrac{-1}{3} \Rightarrow 3x^2 + x + 3 = 0 \Rightarrow x = \dfrac{-1 \pm \sqrt{-35}}{6}$ , this not possible since the question asks for real roots.

When $t = \dfrac{10}{3} , x + \dfrac{1}{x} =\dfrac{10}{3} \Rightarrow 3x^2 - 10x + 3= 0\Rightarrow (3x - 1)(x - 3) = 0$

$\Rightarrow x = \dfrac{1}{3} , 3$

Hence , sum of real values of $x = \dfrac{1}{3} + 3= \dfrac{10}{3}$ [Actually this can be derived easily by Vieta's formula.]

Thus , $m + n = 10 + 3 = \large\boxed{13}$

$Generalization$ : This method which I used is only possible for polynomials of type (known as Symmetric polynomials):

$ax^4 + bx^3 + cx^2 + bx + a$

If $\bigg(x + \dfrac{1}{x}\bigg) = t$ , then such type of polynomials can be factorized as:

$x^2(at^2 + bt + c - 2a)$

I used a graphical approach , Plot the graph and add the zeroes .

One root is 3 and the other comes out to be $\frac{1}{3}$ using Vieta's Rule.

The sum is $3 + \frac{1}{3} = \frac{10}{3} \Rightarrow 13$

Thanks Nihar for such a beautiful generalization . I learnt something new.:-):-):-)

Put the equation as f(x)=0 and use a graphing calculator to find the zeros. After finding the real roots one can solve the remaining quadratic to get the complex roots. (This is not asked for.)

Graphical Analysis is a better approach for this problem. First, rearrange the given equation in the form f(x)=0 (Bring all terms involving x on one side). Then plot f(x) and see where the plot cuts x-axis. They give us information about roots of the equation. We just sum them (Here the roots are x=3 and x=1/3), and get the answer as 13.

P.S. This solution is for Beginners like me trying to solve Level 5 Problems !