Quant Exam > Quant Questions > |x2 + x| – 5 < 0a)x < 0b)x >...
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|x2 + x| – 5 < 0
- a)x < 0
- b)x > 0
- c)All values of x
- d)None of these
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
|x2 + x| – 5 < 0a)x < 0b)x > 0c)All values of xd)Non...
Understanding the Equation
The expression given is |x² + x|, which needs to be compared with the value 5.
Breaking Down the Absolute Value
- The absolute value function |y| is defined as:
- y if y ≥ 0
- -y if y < />
So, we can analyze the expression x² + x in two cases: when it is non-negative and when it is negative.
Case 1: x² + x ≥ 0
- The equation simplifies to:
- x² + x = 5
- Rearranging gives:
- x² + x - 5 = 0
- Applying the quadratic formula:
- x = [−1 ± √(1 + 20)] / 2
- This yields two real solutions.
Case 2: x² + x < />
- Here, the absolute value flips the sign:
- -(x² + x) = 5
- Rearranging gives:
- x² + x + 5 = 0
- The discriminant for this quadratic is negative:
- D = 1 - 20 < />
- Thus, there are no real solutions in this case.
Conclusion
Given that the first case provides two real solutions and the second case yields none, the overall conclusion is that:
- All values of x from the first case are valid solutions.
- Therefore, option **C) All values of x** is indeed correct.
This indicates the expression can take on the value of 5 for various x values, making option C the appropriate choice.
The expression given is |x² + x|, which needs to be compared with the value 5.
Breaking Down the Absolute Value
- The absolute value function |y| is defined as:
- y if y ≥ 0
- -y if y < />
So, we can analyze the expression x² + x in two cases: when it is non-negative and when it is negative.
Case 1: x² + x ≥ 0
- The equation simplifies to:
- x² + x = 5
- Rearranging gives:
- x² + x - 5 = 0
- Applying the quadratic formula:
- x = [−1 ± √(1 + 20)] / 2
- This yields two real solutions.
Case 2: x² + x < />
- Here, the absolute value flips the sign:
- -(x² + x) = 5
- Rearranging gives:
- x² + x + 5 = 0
- The discriminant for this quadratic is negative:
- D = 1 - 20 < />
- Thus, there are no real solutions in this case.
Conclusion
Given that the first case provides two real solutions and the second case yields none, the overall conclusion is that:
- All values of x from the first case are valid solutions.
- Therefore, option **C) All values of x** is indeed correct.
This indicates the expression can take on the value of 5 for various x values, making option C the appropriate choice.
Community Answer
|x2 + x| – 5 < 0a)x < 0b)x > 0c)All values of xd)Non...
At x = 0 inequality is satisfied.
Thus, options (a), (b), and (d) are rejected.
Option (c) is correct.
Thus, options (a), (b), and (d) are rejected.
Option (c) is correct.
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|x2 + x| – 5 < 0a)x < 0b)x > 0c)All values of xd)None of theseCorrect answer is option 'C'. Can you explain this answer?
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