absolute fraction of polynomials

Question

I'm trying to solve inequality:

$$\bigg|\frac{x^2+2x-36}{x^2-4}\bigg|\gt 1$$

The first step I do is to conclude how the function behaves (to see when it's negative):

$$\frac{x^2+2x-36}{x^2-4} = 0$$

My approach is to solve the nominator and denominator separately (as separate second degrees polynomials). The denominator is easy (opens upward parabola): $$x^2-4=0 \\(x-2)(x+2)=0\iff x=-2 \vee x=2 $$ The issue I have is with the nominator: $$x^2+2x-36 = 0$$ Since the standard approach with quadratic formulas doesn't look sensible to me.

Could you share some hints? How I should approach that?

Answer 1

Consider the equality that lies at the boundary of the inequality.

$$\bigg|\frac{x^2+2x-36}{x^2-4}\bigg| = 1$$ $$\frac{x^2+2x-36}{x^2-4} = \pm 1$$ $$x^2+2x-36 = \pm ({x^2-4})$$ $$x^2+2x-36 = x^2-4 \lor x^2+2x-36 = -x^2+4$$ $$2x-32 = 0 \lor 2x^2+2x-40 = 0$$ $$x = 16 \lor (x = 4 \lor x = -5)$$

This gives you three “special” values of $x$ to work with, in addition to the two you already found $x = \pm 2$, where the denominator is 0. In order, these are:

• $x = -5$ (Polynomial fraction is -1.)
• $x = -2$ (Polynomial fraction is undefined.)
• $x = 2$ (Polynomial fraction is undefined.)
• $x = 4$ (Polynomial fraction is -1.)
• $x = 16$ (Polynomial fraction is 1.)

These five values partition the real line into six intervals:

• $x \in (-\infty, -5)$
• $x \in (-5, -2)$
• $x \in (-2, 2)$
• $x \in (2, 4)$
• $x \in (4, 16)$
• $x \in (16, \infty)$

Now, you just have to go through these six cases and determine which ones satisfy your original inequality.

Answer 2

$$ \begin{aligned} & \left|\frac{x^2+2 x-36}{x^2-4}\right|>1, \text { where } x \neq \pm 2 \\ \Rightarrow \quad& \left|x^2+2 x-36\right|>\left|x^2-4\right| \\ \Rightarrow \quad&\left(x^2+2 x-36\right)^2>\left(x^2-4\right)^2 \\ \Rightarrow \quad& \left(x^2+2 x-36\right)^2-\left(x^2-4\right)^2>0 \\ \Rightarrow \quad& \left(x^2+2 x-36+x^2-4\right)\left(x^2+2 x-36-x^2+4\right)>0 \\ \Rightarrow \quad& 4\left(x^2+x-20\right)(x-16)>0 \\ \Rightarrow \quad& 4(x+5)(x-4)(x-16)>0 \\ \Rightarrow \quad& -5<x<4 \text { or } x>16 \end{aligned} $$ However $x\ne \pm 2$, therefore the solutions are $$\boxed{-5<x<-2, \quad-2<x<2, \quad 2<x<4 \quad\textrm{ and }x>16.}$$

Answer 3

Alternative approach:

$$\bigg|\frac{x^2+2x-36}{x^2-4}\bigg|\gt 1$$

This is the approach that I advise for inexperienced Math students:

Let $~N = x^2 + 2x - 36.$
Let $~D = x^2 - 4.$

Note that:

$$\left|\frac{N}{D}\right|= |N| \times \left|\frac{1}{D}\right|.$$

Construe the problem as

$$ |N| \times \left|\frac{1}{D}\right| > 1.$$

Then, to work the problem, I suggest that you:

• Recognize that to algebraically evaluate an expression like $|N|$, you have to identify whether $N$ is positive, zero, or negative.
• Identify those values of $x$ that cause $N$ to be either positive, zero, or negative.
• Identify those values of $x$ that cause $D$ to be either positive, zero, or negative.
• Identify which possibilities need to be manually explored, given the constraint that $~\displaystyle |N| \times \left|\frac{1}{D}\right| > 1.$
• Explore the pertinent cases.
• Collect all of the results into a final answer.

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$\underline{\text{Analysis of} ~N}$

Completing the square, $N = (x+1)^2 - 37.$ So:

• If $(x+1)^2 > 37,~$ then $N > 0.$
• If $(x+1)^2 = 37,~$ then $N = 0.$
• If $(x+1)^2 < 37,~$ then $N < 0.$

At this point, you need the principle that if $~r,s \in \Bbb{R},~$ and $s > 0$, then

• $\displaystyle r^2 < s \iff -\sqrt{s} < r < \sqrt{s}.$
• $\displaystyle r^2 = s \iff r = \pm \sqrt{s}.$
• $\displaystyle r^2 > s \iff \left\{ ~\left[ ~r < -\sqrt{s} ~\right] ~~\text{or}~~ ~\left[ ~r > \sqrt{s} ~\right] ~\right\}.$

Therefore,

• $\displaystyle N > 0 \iff \left\{ ~\left[ ~x+1 < -\sqrt{37} ~\right] ~~\text{or}~~ ~\left[ ~x+1 > \sqrt{37} ~\right] ~\right\}$
  $\displaystyle \iff \left\{ ~\left[ ~x < -\sqrt{37} - 1 ~\right] ~~\text{or}~~ ~\left[ ~x > \sqrt{37} - 1 ~\right] ~\right\}.$

• $\displaystyle N = 0 \iff \left\{ ~\left[ ~x+1 = -\sqrt{37} ~\right] ~~\text{or}~~ ~\left[ ~x+1 = \sqrt{37} ~\right] ~\right\}$
  $\displaystyle \iff \left\{ ~\left[ ~x = -\sqrt{37} - 1 ~\right] ~~\text{or}~~ ~\left[ ~x = \sqrt{37} - 1 ~\right] ~\right\}.$

• $\displaystyle N < 0 \iff -\sqrt{37} < (x + 1) < \sqrt{37}$
  $\displaystyle \iff -\sqrt{37} - 1 < x < \sqrt{37} - 1.$

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$\underline{\text{Analysis of} ~D}$

Similarly

• $\displaystyle D > 0 \iff \left\{ ~\left[ ~x < -2 ~\right] ~~\text{or}~~ ~\left[ ~x > 2 ~\right] ~\right\}.$

• $\displaystyle D = 0 \iff x = \pm 2.$

• $\displaystyle D < 0 \iff -2 < x < 2.$

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$\underline{\text{Identify The Pertinent Cases to Explore}}$

Any values of $x$ that cause $D$ to equal $0$ must be rejected, since you can't divide by $0$.

Any values of $x$ that cause $N$ to equal $0$ must be rejected, because if the overall expression $~\displaystyle \left|\frac{N}{D}\right|~$ is to be greater than $1$, you can't have $N = 0.$

So, you need to consider the following regions:

• For $N$, either
  $~\displaystyle x < -\sqrt{37} -1,$
  or $~~\displaystyle -\sqrt{37} -1 < x < \sqrt{37} -1$
  or $~~\displaystyle \sqrt{37} -1 < x.$

• For $D$, either
  $~\displaystyle x < -2,$
  or $~~\displaystyle -2 < x < 2$
  or $~~\displaystyle 2 < x.$

Combining these considerations, you have the following cases:

• Case 1: $~\displaystyle x < -\sqrt{37} - 1.$

• Case 2: $~\displaystyle -\sqrt{37} - 1 < x < -2.$

• Case 3: $~\displaystyle -2 < x < 2.$

• Case 4: $~\displaystyle 2 < x < \sqrt{37} - 1.$

• Case 5: $~\displaystyle \sqrt{37} - 1 < x.$

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$\underline{\text{Analysis of Case 1:} ~~\displaystyle x < -\sqrt{37} - 1}$

$|N| = |x^2 + 2x - 36| = x^2 + 2x - 36.$

$|D| = |x^2 - 4| = x^2 - 4.$

Therefore, the constraint is

$$\frac{x^2 + 2x - 36}{x^2 - 4} > 1 \iff $$

$$x^2 + 2x - 36 > x^2 - 4 \iff $$

$$2x - 36 > - 4 \iff $$

$$2x > 32 \iff x > 16.$$

Here, there are no values of $x$ that simultaneously satisfy

$$\left[ ~x < -\sqrt{-37} - 1 ~\right] ~~\text{and}~~ \left[ ~x > 16 ~\right].$$

Therefore, Case 1 does not contain any satisfying values of $x$.

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$\underline{\text{Analysis of Case 2:} ~~\displaystyle -\sqrt{37} - 1 < x < -2}$

$|N| = |x^2 + 2x - 36| = 36 - 2x - x^2.$

$|D| = |x^2 - 4| = x^2 - 4.$

Therefore, the constraint is

$$\frac{36 - 2x - x^2}{x^2 - 4} > 1 \iff $$

$$36 - 2x - x^2 > x^2 - 4 \iff $$

$$- 2x^2 - 2x + 40 > 0 \iff $$

$$x^2 + x - 20 < 0 \iff $$

$$(x + 5) \times (x - 4) < 0 \iff $$

$$-5 < x < 4.$$

Combining this with the Case constraint of

$$\displaystyle -\sqrt{-37} - 1 < x < -2$$

yields the satisfying Case 2 values of

$$-5 < x < -2.$$

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$\underline{\text{Analysis of Case 3:} ~~\displaystyle -2 < x < 2}$

$|N| = |x^2 + 2x - 36| = 36 - 2x - x^2.$

$|D| = |x^2 - 4| = 4 - x^2.$

Therefore, the constraint is

$$\frac{36 - 2x - x^2}{4 - x^2} > 1 \iff $$

$$36 - 2x - x^2 > 4 - x^2 \iff $$

$$36 - 2x > 4 \iff x < 16.$$

Combining this with the Case constraint of

$$-2 < x < 2$$

yields the satisfying Case 3 values of

$$-2 < x < 2.$$

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$\underline{\text{Analysis of Case 4:} ~~\displaystyle 2 < x < \sqrt{37} - 1}$

$|N| = |x^2 + 2x - 36| = 36 - 2x - x^2.$

$|D| = |x^2 - 4| = x^2 - 4.$

Therefore, the constraint is

$$\frac{36 - 2x - x^2}{x^2 - 4} > 1 \iff $$

$$36 - 2x - x^2 > x^2 - 4 \iff $$

$$40 - 2x - 2x^2 > 0 \iff $$

$$x^2 + x - 20 < 0 \iff $$

$$(x+5) \times (x-4) < 0 \iff $$

$$-5 < x < 4.$$

Combining this with the Case constraint of

$$2 < x < \sqrt{37} - 1$$

yields the satisfying Case 4 values of

$$2 < x < 4.$$

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$\underline{\text{Analysis of Case 5:} ~~\displaystyle \sqrt{37} - 1 < x}$

$|N| = |x^2 + 2x - 36| = x^2 + 2x - 36.$

$|D| = |x^2 - 4| = x^2 - 4.$

Therefore, the constraint is

$$\frac{x^2 + 2x - 36}{x^2 - 4} > 1 \iff $$

$$x^2 + 2x - 36 > x^2 - 4 \iff $$

$$2x - 36 > - 4 \iff $$

$$2x > 32 \iff 16 < x.$$

Combining this with the Case constraint of

$$\sqrt{37} - 1 < x$$

yields the satisfying Case 5 values of

$$16 < x.$$

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$\underline{\text{Final Answer}}$

• Case 1: No satisfying values.

• Case 2: $~\displaystyle -5 < x < -2.$

• Case 3: $~\displaystyle -2 < x < 2.$

• Case 4: $~\displaystyle 2 < x < 4.$

• Case 5: $~\displaystyle 16 < x.$

So, the mutually exclusive sets represented by Cases 2 through 5 directly above identify all of the satisfying values of $x$.