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30 Nov 2021, 04:19
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A
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Difficulty:
55%
(hard)
Question Stats:
64% (01:55) correct
36%
(02:20)
wrong
based on 11
sessions
History
Date
Time
Result
Not Attempted Yet
\(|x + 5|x|| < 12\)
What is the sum of all integers which satisfy the inequality above?
A. \(-2\)
B. \(-1\)
C. \(0\)
D. \(1\)
E. \(2\)
What is the sum of all integers which satisfy the inequality above?
A. \(-2\)
B. \(-1\)
C. \(0\)
D. \(1\)
E. \(2\)
30 Nov 2021, 04:19
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Official Solution:
\(|x + 5|x|| < 12\)
What is the sum of all integers which satisfy the inequality above?
A. \(-2\)
B. \(-1\)
C. \(0\)
D. \(1\)
E. \(2\)
\(|x + 5|x|| < 12\)
\(-12 < x + 5|x| < 12\)
CASE 1: \(x < 0\).
\(-12 < x + 5(-x) < 12\)
\(-12 < -4x < 12\)
Multiply by \(-\frac{1}{4}\) (and don't forget to flip the signs since we multiply by a negative number): \(3 > x > -3\)
Negative integers in this range (don't forget that we consider \(x < 0\) for this case) are -1, and -2.
CASE 2: \(x \geq 0\).
\(-12 < x + 5x < 12\)
\(-12 < 6x < 12\)
Divide by \(6\): \(-2 < x < 2\)
Non-negative integers in this range (don't forget that we consider \(x \geq 0\) for this case) are 0, and 1.
The sum of all integers which satisfy the inequality is therefore: \(-1+(-2)+0+1=-2\)
Answer: A
\(|x + 5|x|| < 12\)
What is the sum of all integers which satisfy the inequality above?
A. \(-2\)
B. \(-1\)
C. \(0\)
D. \(1\)
E. \(2\)
\(|x + 5|x|| < 12\)
\(-12 < x + 5|x| < 12\)
CASE 1: \(x < 0\).
\(-12 < x + 5(-x) < 12\)
\(-12 < -4x < 12\)
Multiply by \(-\frac{1}{4}\) (and don't forget to flip the signs since we multiply by a negative number): \(3 > x > -3\)
Negative integers in this range (don't forget that we consider \(x < 0\) for this case) are -1, and -2.
CASE 2: \(x \geq 0\).
\(-12 < x + 5x < 12\)
\(-12 < 6x < 12\)
Divide by \(6\): \(-2 < x < 2\)
Non-negative integers in this range (don't forget that we consider \(x \geq 0\) for this case) are 0, and 1.
The sum of all integers which satisfy the inequality is therefore: \(-1+(-2)+0+1=-2\)
Answer: A
01 Apr 2025, 05:57
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|x+5|x||<12
Can be opened to equations:
x + 5x < 12 => 6x < 12 => x < 2
x - 5x < 12 => -4x < 12 => -x < 3 => x > -3
-x -5x < 12 => -6x < 12 => -x < 2 => x > -2
-x + 5x < 12 => 4x < 12 => x < 3
From the above 4 ranges of x we can say that -2 < x < 2 so integer values are -1, 0 and 1 which give the sum as 0.
I understand that putting x = -2 satisfies the equation, but where am I wrong in my approach?
Bunuel can you help please?
Can be opened to equations:
x + 5x < 12 => 6x < 12 => x < 2
x - 5x < 12 => -4x < 12 => -x < 3 => x > -3
-x -5x < 12 => -6x < 12 => -x < 2 => x > -2
-x + 5x < 12 => 4x < 12 => x < 3
From the above 4 ranges of x we can say that -2 < x < 2 so integer values are -1, 0 and 1 which give the sum as 0.
I understand that putting x = -2 satisfies the equation, but where am I wrong in my approach?
Bunuel can you help please?
