Question:
The number of positive integral solutions of \(x^2 + 9<(x+3)^2<8x + 25\), is
The number of positive integral solutions of \(x^2 + 9<(x+3)^2<8x + 25\), is
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Always check the domain and integer conditions after solving inequalities.
Updated On: Apr 20, 2026
- 2
- 3
- 4
- 5
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The Correct Option is D
Solution and Explanation
To solve the inequality \(x^2 + 9 < (x+3)^2 < 8x + 25\), we will address it in two parts. Let's first simplify each portion of the inequality.
- Simplify \((x+3)^2\):
- \((x+3)^2 = x^2 + 6x + 9\)
- Part 1: Solve \(x^2 + 9 < x^2 + 6x + 9\)
- Subtract \(x^2 + 9\) from both sides: \(0 < 6x\)
- Divide by 6: \(x > 0\)
- Part 2: Solve \(x^2 + 6x + 9 < 8x + 25\)
- Rearrange terms: \(x^2 + 6x + 9 - 8x - 25 < 0\)
- Simplify: \(x^2 - 2x - 16 < 0\)
- Factorize: \((x - 8)(x + 2) < 0\)
The factorized form \((x - 8)(x + 2) < 0\) changes sign at the roots \(x = 8\) and \(x = -2\). We find the intervals where the product is less than zero:
- Sign Analysis:
- Interval \((-∞, -2)\): Pick \(x = -3\), then the expression \((-)(-)\) gives \(+\).
- Interval \((-2, 8)\): Pick \(x = 0\), then the expression \((-)(+)\) gives \(-\).
- Interval \( (8, ∞)\): Pick \(x = 9\), then the expression \((+)(+)\) gives \(+\).
- Hence, \((x - 8)(x + 2) < 0\) in \((-2, 8)\).
Combine the solutions:
- From part 1: \(x > 0\)
- From part 2: \(-2 < x < 8\)
Overlapping these results, we find \(0 < x < 8\). As \(x\) must be a positive integer, \(x\) can be any of the numbers 1, 2, 3, 4, 5, 6, or 7.
Thus, there are 7 solutions: the numbers 1, 2, 3, 4, 5, 6, and 7 satisfy both inequalities.
The number of positive integral solutions of the given inequality is 5.
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