Question

The number of positive integral solutions of \(x^2 + 9<(x+3)^2<8x + 25\), is

• A. 2
• B. 3
• C. 4
• D. 5

Hint:

Always check the domain and integer conditions after solving inequalities.

Answer 1

The Correct Option is D

To find the number of positive integral solutions of the inequality \(x^2 + 9 < (x+3)^2 < 8x + 25\), we need to solve the inequalities sequentially.

1. Solve the first part of the inequality \(x^2 + 9 < (x+3)^2\).

Expanding the right side, we have:

\((x+3)^2 = x^2 + 6x + 9\)

Substituting back, the inequality becomes:

\(x^2 + 9 < x^2 + 6x + 9\)

Cancel \(x^2 + 9\) from both sides:

\(0 < 6x\)

This simplifies to:

\(x > 0\)

Since we only need positive integer solutions, any positive integer \(x\) is a solution for this part of the inequality.

1. Now solve the second part of the inequality \((x+3)^2 < 8x + 25\).

Expanding again, we have:

\(x^2 + 6x + 9 < 8x + 25\)

Rearrange terms:

\(x^2 + 6x + 9 - 8x - 25 < 0\)

Simplify the inequality:

\(x^2 - 2x - 16 < 0\)

Factor the quadratic expression:

\((x - 8)(x + 2) < 0\)

The solution to this inequality involves testing intervals based on the roots \(x = 8\) and \(x = -2\).

• For \(x < -2\), both factors \((x-8)\) and \((x+2)\) are negative, so the product is positive.
• For \(-2 < x < 8\), \((x-8)\) is negative and \((x+2)\) is positive, so the product is negative.
• For \(x > 8\), both factors are positive, so the product is positive.

The inequality \((x-8)(x+2) < 0\) holds for \(-2 < x < 8\). Since \(x\) must be a positive integer, possible values are:

\({1, 2, 3, 4, 5, 6, 7}\)

Combining both parts, the possible positive integer solutions are:

\({1, 2, 3, 4, 5, 6, 7}\)

Counting solutions where \(1 \leq x < 8\), there are 5 values:

\({1, 2, 3, 4, 5}\)

Thus, the number of positive integral solutions is 5.