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# How many real numbers x satisfy the inequality below?

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Manager
Joined: 21 Feb 2019
Posts: 114
Location: Italy
How many real numbers x satisfy the inequality below?  [#permalink]

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01 Apr 2019, 06:47
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6
00:00

Difficulty:

95% (hard)

Question Stats:

22% (02:06) correct 78% (02:05) wrong based on 50 sessions

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How many real numbers x satisfy the inequality below?

$$|x^4 -4x^2 -6| ≥ |x^4 -4x^2 +14|$$

A. 0
B. 1
C. 2
D. 4
E. Infinitely many

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Joined: 21 Feb 2019
Posts: 114
Location: Italy
Re: How many real numbers x satisfy the inequality below?  [#permalink]

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01 Apr 2019, 16:36
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IanStewart I really appreciated the reasoning behind your conclusion (focus on modulus meaning).

More straightforward and less intuitive:

$$x^4 - 4x^2 = t$$

$$|t - 6| ≥ |t + 14|$$

$$(t - 6)^2 ≥ (t + 14)^2$$

$$t^2 +36 -12t ≥ t^2 +196 +28t$$

$$40t ≤ - 160$$

$$t ≤ - 4$$

Substitute $$t$$:

$$x^4 - 4x^2 +4 ≤ 0$$

$$(x^2 - 2)^2 ≤ 0$$

$$x = ± \sqrt{2}$$
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##### General Discussion
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Re: How many real numbers x satisfy the inequality below?  [#permalink]

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01 Apr 2019, 13:35
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Let y = x^4 - 4x^2. Then the inequality becomes

|y - 6| > |y + 14|

|y - 6| > |y - (-14)|

Since |a-b| is the distance on the number line between a and b, this inequality just says, in words, "the distance between y and 6 is greater than or equal to the distance between y and -14", or in other words, "y is closer to -14 than it is to 6". So y is at or less than the midpoint between 6 and -14, and y must be less than or equal to -4. Since y = x^4 - 4x^2, we just need to determine when:

x^4 - 4x^2 < -4
x^4 - 4x^2 + 4 < 0
(x^2 - 2)^2 < 0

and since the left side above is a square, it cannot be negative, so we only have solutions when (x^2 - 2)^2 = 0, so when x^2 - 2 = 0, and x^2 = 2. That equation has two solutions, x = √2 and x = -√2, so the answer is C.
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Joined: 07 Apr 2019
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How many real numbers x satisfy the inequality below?  [#permalink]

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08 Apr 2019, 19:28
lucajava wrote:
How many real numbers x satisfy the inequality below?

$$(I) \quad |x^4 -4x^2 -6| ≥ |x^4 -4x^2 +14|$$

A. 0 B. 1 C. 2 D. 4 E. Infinitely many

The substitution $$z = x^2$$ brings us into the realm of the more familiar quadratic functions.

$$\lvert z^2 -4z -6\rvert ≥ \lvert z^2 -4z +14\rvert$$

The two parabolas $$p_1\colon y = z^2 -4z -6$$ and $$p_2\colon y = z^2 -4z +14$$ share the same line of symmetry $$z = 2$$, and $$p_1$$ is actually just $$p_2$$ shifted downward. Their vertices are $$(2,-10)$$ and $$(2,+10)$$, respectively, and $$p_2(z)$$ is strictly positive (you could simply drop the absolute value). Flipping the negative part of $$p_1$$ over the $$z$$-axis results in a kissing of $$p_2$$ at $$z = 2$$, but otherwise the flipped $$p_1$$ stays strictly below $$p_2$$. Therefore, the inequality (I) can only be true for $$z = x^2 = 2$$, i.e. $$x = \pm \sqrt{2}$$.
How many real numbers x satisfy the inequality below?   [#permalink] 08 Apr 2019, 19:28
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