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How many real numbers x satisfy the inequality below?

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How many real numbers x satisfy the inequality below?  [#permalink]

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New post 01 Apr 2019, 06:47
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A
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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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Re: How many real numbers x satisfy the inequality below?  [#permalink]

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New post 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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Re: How many real numbers x satisfy the inequality below?  [#permalink]

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New post 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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How many real numbers x satisfy the inequality below?  [#permalink]

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New post 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. \(x = \pm \sqrt{2}\).
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How many real numbers x satisfy the inequality below?   [#permalink] 08 Apr 2019, 19:28
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