5
$\begingroup$

I want to solve the following equation for $x$

$$\left(x + \frac{6}{x} \right)^2 + \left( x + \frac{6}{x} \right) = 30$$

I done my working till -

$$x^4 + x^3 - 18x^2 + 6x + 36 = 0$$

From here how do I solve for $x$ when I have any different powers ?

$\endgroup$
3
  • 12
    $\begingroup$ Why not start with $y=x+{6\over x}$ and solve for $y$ first? $\endgroup$ – abiessu Jun 21 '16 at 16:39
  • $\begingroup$ Or factor the equation you end up with $(x-3)(x-2)(x^2+6x+6)$. $\endgroup$ – almagest Jun 21 '16 at 16:40
  • 1
    $\begingroup$ Note that the equation you started with has a nice structure. The "simplified" quartic, not so much. $\endgroup$ – André Nicolas Jun 21 '16 at 16:45
8
$\begingroup$

Hint: set $$t=x+\frac{6}{x}$$ and you will get a quadratic equation in $t$

$\endgroup$
0
$\begingroup$

$t^2+t=30$ iff $t=5$ or $t=-6$, and $$ x+\frac{6}{x}\in\{-6,5\} $$ iff $\color{red}{x\in\{2,3,-3-\sqrt{3},-3+\sqrt{3}\}}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.