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How do I solve for $x$ algebraically?

$$\dfrac{x^2(x^2-1)}{x+3} = 12$$

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Multiple both sides by $x+3$ and solve the resulting quartic. –  copper.hat Oct 20 '12 at 17:31
how do I solve that quartic? I don't want a guess-and-check or calculator solution. –  makosman12 Oct 20 '12 at 17:33
and $x \neq -3 $. –  Iuli Oct 20 '12 at 17:33
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4 Answers

up vote 2 down vote accepted

Hint: Rewrite as $x^4-x^2-12x-36=0$. Note that by great good "luck" $x=-2$ and $x=3$ are solutions. (We crossed our fingers and hoped for rational solutions. By the Rational Root Theorem, these have to be of the form $p/q$ where $p$ divides $-36$ and $q$ divides $1$.) If we had not been lucky, we would be faced with a "general" quartic. Finding solutions to a quartic is in general possible, but very painful.

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Damn, you are fast. –  copper.hat Oct 20 '12 at 17:35
What if I don't want luck? How do I do it algebraically? –  makosman12 Oct 20 '12 at 17:35
Look at the general solution of the quartic, which goes back to Cardano and Ferrari ($16$th century, not the car Ferrari). It is messy. After the two solutions that I gave, you still need to divide the quartic by $(x+2)9x-3)$ to obtain a quadratic. –  André Nicolas Oct 20 '12 at 17:40
All right thanks –  makosman12 Oct 20 '12 at 17:47
Note that the Rational Root Theorem is a bit of algebra, or number theory, that we used in our search. The Cardano-Ferrari solution, and variants found by Descartes, Newton, Lagrange, and others are all usually quite impractical. One can find expressions for the roots in terms of square roots and cube roots. But these are typically square roots and cube roots of complex numbers. One can use trigonometric methods to get at these, but they probably do not qualify as algebra. –  André Nicolas Oct 20 '12 at 17:51
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Note that the equation you get after multiplying out can be rearranged as $$x^4-(x^2+12x+36) = x^4-(x+6)^2 = (x^2+x+6)(x^2-x-6) = 0$$

The second factor gives the roots $x=3$ and $x=-2$ - without guessing. You have to spot the form, though.

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Multiplying both sides by $x+3$ gives $x^2(x^2-1) = 12 (x+3)$. This can be re-written as $x^4-x^2-12x-36 = 0$. A few guesses shows that $3$ and $-2$ are solutions, so you can factor these out to get a quadratic.

Explicitly, the equation becomes $(x-3)(x+2)(x^2+x+6) = 0$, hence the solutions are $3$, $-2$, and $\frac{1}{2}(-1 \pm i \sqrt{23}$).

(And, of course, we check that none of these are $-3$.)

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Is there no solution to such a polynomial that involves no guessing? –  makosman12 Oct 20 '12 at 17:36
There is a formula, but it is messy. Wiki is your friend :-). –  copper.hat Oct 20 '12 at 17:39
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Rearranging the equation, we get $$x^2(x^2-1) = 12(x+3)$$ $$x^4 - x^2 - 12x - 36 = 0$$ First we search for integer roots. The integer root must divide $36$. Hence, the possible integer options are $\pm 1,\pm 2, \pm 3$. Checking these $6$ options, give us $x=-2$ and $x=3$. Hence, $$x^4 - x^2 - 12x - 36 = (x+2)(x-3)(x^2+ax+b)$$ Comparing coefficients, we get $a=1$ and $b=6$. Solving the quadratic, gives the other roots as $$x^2 + x + 6 =0 \implies \left(x + \dfrac12 \right)^2 + 6 - \dfrac14 = 0 \implies x = -\dfrac12 \pm i \dfrac{\sqrt{23}}2$$

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