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

Options

A. 2
B. 3
C. 4
D. 5

Explanation

To solve the inequality \( x^2 + 9 < (x + 3)^2 < 8x + 25 \) and find the number of integral solutions, we will break it down into two parts.

### Step 1: Solve the first inequality \( x^2 + 9 < (x + 3)^2 \)

1. Expand the right side:
\[ (x + 3)^2 = x^2 + 6x + 9 \]
So, the inequality becomes:
\[ x^2 + 9 < x^2 + 6x + 9 \]

2. Simplify the inequality:
\[ x^2 + 9 - x^2 - 9 < 6x \]
This simplifies to:
\[ 0 < 6x \]
or
\[ x > 0 \]

### Step 2: Solve the second inequality \( (x + 3)^2 < 8x + 25 \)

1. Again, expand the left side:
\[ (x + 3)^2 = x^2 + 6x + 9 \]
So, the inequality becomes:
\[ x^2 + 6x + 9 < 8x + 25 \]

2. Rearrange the inequality:
\[ x^2 + 6x + 9 - 8x - 25 < 0 \]
This simplifies to:
\[ x^2 - 2x - 16 < 0 \]

3. Factor the quadratic:
\[ x^2 - 2x - 16 = 0 \]
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-16)}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 + 64}}{2} = \frac{2 \pm \sqrt{68}}{2} = \frac{2 \pm 2\sqrt{17}}{2} = 1 \pm \sqrt{17} \]

4. The roots are \( 1 - \sqrt{17} \) and \( 1 + \sqrt{17} \). The quadratic opens upwards, so the solution to the inequality \( x^2 - 2x - 16 < 0 \) is:
\[ 1 - \sqrt{17} < x < 1 + \sqrt{17} \]

### Step 3: Combine the results

1. From the first inequality, we have \( x > 0 \).
2. From the second inequality, we have \( 1 - \sqrt{17} < x < 1 + \sqrt{17} \).

Since \( \sqrt{17} \) is approximately \( 4.123 \), we can calculate:
- \( 1 - \sqrt{17} \approx 1 - 4.123 \approx -3.123 \) (not relevant since \( x > 0 \))
- \( 1 + \sqrt{17} \approx 1 + 4.123 \approx 5.123 \)

Thus, the combined range for \( x \) is:
\[ 0 < x < 5.123 \]

### Step 4: Find the integral solutions

The integral solutions in the range \( 0 < x < 5.123 \) are:
- \( x = 1, 2, 3, 4, 5 \)

### Conclusion

The number of integral solutions is \( 5 \).