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23 Apr 2024, 01:30
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B
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Difficulty:
25%
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Question Stats:
79% (02:11) correct
21%
(02:18)
wrong
based on 368
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If \(x\) is a non-zero integer and \(|6- 3x| +3|x + 2| = 12\), what is the value of \(x^4 - 5x^2 + 4\)?
A. -12
B. 0
C. 4
D. 6
E. 12
A. -12
B. 0
C. 4
D. 6
E. 12
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04 May 2024, 20:45
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Official Solution:
If \(x\) is a non-zero integer and \(|6- 3x| +3|x + 2| = 12\), what is the value of \(x^4 - 5x^2 + 4\)?
A. -12
B. 0
C. 4
D. 6
E. 12
Reduce \(|6 - 3x| + 3|x + 2| = 12\) by 3 to get \(|x - 2| + |x + 2| = 4\).
Notice that if \(x\) is less than -2 or greater than 2, then \(|x - 2| + |x + 2|\) will be more than 4. For example, if \(x < -2\), then \(|x - 2| > 4\), and if \(x > 2\), then \(|x + 2| > 4\). Thus, \(x\) must be between -2 and 2, inclusive.
Next, when \(-2 \leq x \leq 2\), \(|x - 2| = -(x-2)\) and \(|x + 2| = x+2\). Therefore, \(|x - 2| + |x + 2| = 4\) simplifies to \(-(x - 2) + (x + 2) = 4\), which simplifies further to \(4 = 4\). This implies that ANY value of \(x\) such that \(-2 \leq x \leq 2\) satisfies \(|x - 2| + |x + 2| = 4\).
We are told that \(x\) is a non-zero integer; thus, \(x\) can only be -2, -1, 1, or 2. For each of these values, \(x^4 - 5x^2 + 4\) equals 0.
Answer: B
If \(x\) is a non-zero integer and \(|6- 3x| +3|x + 2| = 12\), what is the value of \(x^4 - 5x^2 + 4\)?
A. -12
B. 0
C. 4
D. 6
E. 12
Reduce \(|6 - 3x| + 3|x + 2| = 12\) by 3 to get \(|x - 2| + |x + 2| = 4\).
Notice that if \(x\) is less than -2 or greater than 2, then \(|x - 2| + |x + 2|\) will be more than 4. For example, if \(x < -2\), then \(|x - 2| > 4\), and if \(x > 2\), then \(|x + 2| > 4\). Thus, \(x\) must be between -2 and 2, inclusive.
Next, when \(-2 \leq x \leq 2\), \(|x - 2| = -(x-2)\) and \(|x + 2| = x+2\). Therefore, \(|x - 2| + |x + 2| = 4\) simplifies to \(-(x - 2) + (x + 2) = 4\), which simplifies further to \(4 = 4\). This implies that ANY value of \(x\) such that \(-2 \leq x \leq 2\) satisfies \(|x - 2| + |x + 2| = 4\).
We are told that \(x\) is a non-zero integer; thus, \(x\) can only be -2, -1, 1, or 2. For each of these values, \(x^4 - 5x^2 + 4\) equals 0.
Answer: B
General Discussion
23 Apr 2024, 02:35
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Bunuel
The critical points are -2 and 2.
Let's plot these over the number line
---------- Region 1 --------- -2 ---------- Region 2 --------- 2 ---------- Region 3 ---------
In Region 1 (\(x < -2\))
\(|6- 3x| +3|x + 2| = 12\)
\((6- 3x) -3(x + 2) = 12\)
\(6 - 3x -3x - 6 = 12\)
\(x = -2\) → No valid point
In Region 2 \((-2 \leq x \leq 2)\)
\(|6- 3x| +3|x + 2| = 12\)
\((6- 3x) +3(x + 2) = 12\)
\(6 - 3x -3x + 6 = 12\)
\(12 = 12\) → x = {-2, -1, 1, 2}
In Region 3 (\(x > 2\))
\(|6- 3x| +3|x + 2| = 12\)
\(-6 + 3x + 3x + 6= 12\)
\(x = 2 \)
\(x = 2\) → No valid point
Hence, \(x\) can be {-2, -1 , 1, 2} → For all values of x, \(x^4 - 5x^2 + 4\) = 0
Option B
