Absolute Value Inequality
Algebra
Level
3
The set of real solutions to 5|x|≤2|x+1| can be expressed as −A divided by B ≤ x ≤ C dividedby D, where A and B are relatively prime positive integers, and C and D are relatively prime positive integers. Find A+B+C+D.
The answer is 14.
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1 solution
Since |x|2=x2, we can square both sides of the inequality to get 25x2≤4(x+1)2. (Also, 5|x| and 2|x+1| are nonnegative, so this inequality is equivalent to the original inequality.) This equation simplifies to 21x2−8x−4≤0, which factors as (7x+2)(3x−2)≤0. The solution to this inequality is −2 BY 7≤ x ≤2 BY 3.