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 ?
$\begingroup$
$\endgroup$
3
-
12$\begingroup$ Why not start with $y=x+{6\over x}$ and solve for $y$ first? $\endgroup$– abiessuJun 21, 2016 at 16:39
-
$\begingroup$ Or factor the equation you end up with $(x-3)(x-2)(x^2+6x+6)$. $\endgroup$– almagestJun 21, 2016 at 16:40
-
1$\begingroup$ Note that the equation you started with has a nice structure. The "simplified" quartic, not so much. $\endgroup$– André NicolasJun 21, 2016 at 16:45
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
Hint: set $$t=x+\frac{6}{x}$$ and you will get a quadratic equation in $t$
answered Jun 21, 2016 at 16:40
$\begingroup$
$\endgroup$
$t^2+t=30$ iff $t=5$ or $t=-6$, and $$ x+\frac{6}{x}\in\{-6,5\} $$ iff $\color{red}{x\in\{2,3,-3-\sqrt{3},-3+\sqrt{3}\}}$.
answered Jun 21, 2016 at 18:09