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Solve $$\dfrac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=0.$$ We have $D_x:\begin{cases}x^2-5x+4\ge0\\x^2-5x+4\ne0\end{cases}\iff x^2-5x+4>0\iff x\in(-\infty;1)\cup(4;+\infty).$ Now I am trying to solve the equation $x^3-4x^2-4x+16=0.$ I have not studied how to solve cubic equations. Thank you in advance!

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7 Answers 7

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It's $$x^2(x-4)-4(x-4)=0$$ or $$(x-4)(x^2-4)=0.$$ Can you end it now?

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We have that

$$x^3-4x^2-4x+16=x(x^2-4x+4)-8x+16=x(x-2)^2-8(x-2)=$$

$$=(x-2)(x^2-2x-8)=(x-2)(x+2)(x-4)=0$$

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$$\frac{x^3-4x^2-4x+16}{\sqrt{x^2-5x+4}}=\frac{(x-4)(x-2)(x+2)}{\sqrt{(x-4)(x-1)}}==\frac{\sqrt{x-4}(x-2)(x+2)}{\sqrt{x-1}}$$ The fraction will be $=0$ if the numerator is $0$ and the denominator is not $0$. That is true for $x=2,-2,4$.

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    $\begingroup$ In this case it is still zero even at $x=4$ though, the denominator is not a power one but rather one half ! $\endgroup$
    – Anthony
    Oct 9, 2020 at 19:22
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Use the fact that\begin{align}x^3-4x^2-4x+16=0&\iff x(x^2-4)-4(x^2-4)=0\\&\iff(x-4)(x^2-4)=0.\end{align}

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The other answers pretty much covered all the aspects, I'm just adding an heuristic way of obtaining integer solutions of polynomial equations that can come handy sometimes. The equation $x^3-4x^2-4x+16=0$ can be rewritten as $$ x(x^2-4x-4)= -16 $$ (it's just sending the constant term to the rhs and factoring $x$ on the lhs)

So you see that any integer solution must divide 16. Since the divisors of 16 are $\pm 2, \pm 4, \pm 8, \pm 16$ (and $\pm 1$, if you will), if there are any integer solutions, they must be in the set $\{-16,-8,\cdots, 8, 16\}$. In this case, trying solutions in this set will yield all three solutions to the equation.

Naturally, if there are no integer solutions, this gets you nowhere.

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  • $\begingroup$ Thank you for the response! I don't understand how from $x(x^2-4x-4)=-16$ we see that any integer solution must divide $16$. Can you clarify that for me? Thank you in advance! $\endgroup$ Oct 10, 2020 at 10:43
  • $\begingroup$ @LYI Assuming integer solutions, both $x$ and $x^2-4x-4$ are integers. If their product is $-16$, both those factors must divide 16. $\endgroup$ Oct 10, 2020 at 17:09
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Alternate approach

My algebra abilities are limited. I will show you how I would attack the problem.

Given $$\frac{f(x)}{\sqrt{g(x)}} = 0$$

where (if I understand correctly) $x$ may be any real number

then my first step is to automatically convert the equation to

$$\frac{[f(x)]^2}{g(x)} = 0$$

with the understanding that any values of $x$ that satisfy the second equation must be manually examined to see if they also satisfy the first equation.

My next step, which I consider mandatory in this problem is to meta-cheat.

It can be presumed that you would not have been given this problem unless a solution could be arrived at through the reasonable use of the tools that you have been offered in your class.

Furthermore, attacking cubic equations (let alone $6^{\text{th}}$ degree equations) through brute force is generally considered off limits, especially if you have not been studying cubic equations in class.

At this point, there are only two possiblities:

  1. The teacher or book author is not of sound mind.

  2. There is some hidden elegance that you are expected to discover.

At this point, the only possible elegance that I can imagine (which allows for the possibility that my imagination is too narrow) consists of seeing if $f(x)$ and $g(x)$ can be factored without much trouble. If so, then you can remove common factors, which will simplify the examination of

$$\frac{[f(x)]^2}{g(x)}.$$

There are two ways to handle this. One way is to notice that $g(x) = (x-1)(x-4)$ and then ask yourself whether either of those two factors is also a factor of $f(x)$.

The other alternative, given that you are not supposed to use brute force against a cubic, is to accidentally notice that the coefficients of $f(x)$ are

$$ 1, -4, -4, 16$$

This suggests in and of itself that $(x-4)$ might be a factor of $f(x)$.

However you determine the common factor(s), and simplify the problem, at this point the meta-cheating is concluded, and you can then attack the simplified problem more easily.

By the way
Once you factor $g(x) = (x-1)(x-4)$
you must then immediately presume that
neither $x=1$ or $x=4$ can be considered as satisfactory answers.

This is because either of those two values for $x$ would cause the denominator in the original problem to $= 0$, which is forbidden.

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Answer : $\frac{x^3-4x^2-4x+16}{\sqrt{x^2 - 5x+4}}= \frac{x(x^2 - 4)-4(x^2 - 4)}{\sqrt{x^2 - 5x+4}}$=$\frac{(x^2 - 4)(x-4)}{\sqrt{x^2 - 5x+4}}$

$\sqrt{x^2 - 5x+4} = 0 $ if $ x =(1, 4) $

Suppose $x≠(1,4)$

$(x^2 - 4)(x-4)$ =0

$\Rightarrow$ $ (x - 2)(x+2)(x-4)=0$

$\Rightarrow $ the solution is ($x=2$ or $ x=- 2)$

Because if $x =4 $the denominator equal $0$ So $x=4$ not a solution

Finally : $S=(2,-2) $

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