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If |x - 3| + |x + 9| = 14 then what is the difference between maximum and minimum possible absolute values of x?
- a)4
- b)14
- c)6
- d)0
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
If |x - 3| + |x + 9| = 14 then what is the difference between maximum...
To solve this problem, we need to consider the different cases for the absolute values of x - 3 and x + 9.
Case 1: x - 3 ≥ 0 and x + 9 ≥ 0
In this case, both absolute values are positive. Therefore, we can write the equation as follows:
(x - 3)(x + 9) = 14
Expanding the equation, we get:
x^2 + 9x - 3x - 27 = 14
x^2 + 6x - 41 = 0
Using the quadratic formula, we can solve for x:
x = (-6 ± √(6^2 - 4(1)(-41))) / 2(1)
x = (-6 ± √(36 + 164)) / 2
x = (-6 ± √200) / 2
x = (-6 ± 2√50) / 2
x = -3 ± √50
The minimum possible value of x in this case is -3 - √50, and the maximum possible value is -3 + √50.
Case 2: x - 3 < 0="" and="" x="" +="" 9="" ≥="" />
In this case, the absolute value of x - 3 is negative, and the absolute value of x + 9 is positive. Therefore, we can write the equation as follows:
-(x - 3)(x + 9) = 14
Expanding the equation and simplifying, we get:
x^2 + 9x - 3x - 27 = -14
x^2 + 6x - 13 = 0
Using the quadratic formula, we can solve for x:
x = (-6 ± √(6^2 - 4(1)(-13))) / 2(1)
x = (-6 ± √(36 + 52)) / 2
x = (-6 ± √88) / 2
x = (-6 ± 2√22) / 2
x = -3 ± √22
The minimum possible value of x in this case is -3 - √22, and the maximum possible value is -3 + √22.
Case 3: x - 3 ≥ 0 and x + 9 < />
In this case, the absolute value of x - 3 is positive, and the absolute value of x + 9 is negative. Therefore, the equation is not possible as the product of two opposite signs cannot be equal to 14.
Case 4: x - 3 < 0="" and="" x="" +="" 9="" />< />
In this case, both absolute values are negative. Therefore, we can write the equation as follows:
-(x - 3)(x + 9) = 14
Expanding the equation and simplifying, we get:
x^2 + 9x - 3x - 27 = -14
x^2 + 6x - 13 = 0
Using the quadratic formula, we can solve for x:
x = (-6 ± √(6^2 - 4(1)(-13))) / 2(1)
x = (-6 ± √(36 + 52)) / 2
Case 1: x - 3 ≥ 0 and x + 9 ≥ 0
In this case, both absolute values are positive. Therefore, we can write the equation as follows:
(x - 3)(x + 9) = 14
Expanding the equation, we get:
x^2 + 9x - 3x - 27 = 14
x^2 + 6x - 41 = 0
Using the quadratic formula, we can solve for x:
x = (-6 ± √(6^2 - 4(1)(-41))) / 2(1)
x = (-6 ± √(36 + 164)) / 2
x = (-6 ± √200) / 2
x = (-6 ± 2√50) / 2
x = -3 ± √50
The minimum possible value of x in this case is -3 - √50, and the maximum possible value is -3 + √50.
Case 2: x - 3 < 0="" and="" x="" +="" 9="" ≥="" />
In this case, the absolute value of x - 3 is negative, and the absolute value of x + 9 is positive. Therefore, we can write the equation as follows:
-(x - 3)(x + 9) = 14
Expanding the equation and simplifying, we get:
x^2 + 9x - 3x - 27 = -14
x^2 + 6x - 13 = 0
Using the quadratic formula, we can solve for x:
x = (-6 ± √(6^2 - 4(1)(-13))) / 2(1)
x = (-6 ± √(36 + 52)) / 2
x = (-6 ± √88) / 2
x = (-6 ± 2√22) / 2
x = -3 ± √22
The minimum possible value of x in this case is -3 - √22, and the maximum possible value is -3 + √22.
Case 3: x - 3 ≥ 0 and x + 9 < />
In this case, the absolute value of x - 3 is positive, and the absolute value of x + 9 is negative. Therefore, the equation is not possible as the product of two opposite signs cannot be equal to 14.
Case 4: x - 3 < 0="" and="" x="" +="" 9="" />< />
In this case, both absolute values are negative. Therefore, we can write the equation as follows:
-(x - 3)(x + 9) = 14
Expanding the equation and simplifying, we get:
x^2 + 9x - 3x - 27 = -14
x^2 + 6x - 13 = 0
Using the quadratic formula, we can solve for x:
x = (-6 ± √(6^2 - 4(1)(-13))) / 2(1)
x = (-6 ± √(36 + 52)) / 2
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Community Answer
If |x - 3| + |x + 9| = 14 then what is the difference between maximum...
Case I: x≥3
∴ (x−3)+(x+9)=14
∴ x=4
Case II: 3≥x≥−9
∴ (3−x)+(x+9)=14
This case is not possible.
Case III: x≤−9
∴(3−x)−(x+9)=14
∴ x=−10
The difference between the maximum and the minimum absolute values of x = 10 - 4 = 6, Hence option 3.
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