The number of the integer solutions of `x^(2)+9lt(x+3)^(2)lt8x+25` is

Options

A. one
B. two
C. three
D. none

Answer

To find the number of integer solutions for the inequality \( x^2 + 9 < (x + 3)^2 < 8x + 25 \), we can break it down into two parts:

1. Solve the inequality \( x^2 + 9 < (x + 3)^2 \)
2. Solve the inequality \( (x + 3)^2 < 8x + 25 \)

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

Start by expanding the right side:

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

Now, substitute this back into the inequality:

\[ x^2 + 9 < x^2 + 6x + 9 \]

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

\[ 0 < 6x \]

This simplifies to:

\[ x > 0 \]

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

Again, expand the left side:

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

Substituting back gives us:

\[ x^2 + 6x + 9 < 8x + 25 \]

Rearranging this inequality:

\[ x^2 + 6x + 9 - 8x - 25 < 0 \]

This simplifies to:

\[ x^2 - 2x - 16 < 0 \]

### Step 3: Factor the quadratic inequality

To factor \( x^2 - 2x - 16 \):

\[ x^2 - 2x - 16 = (x - 4)(x + 4) \]

Now we need to find the intervals where this product is less than zero. The critical points are \( x = -4 \) and \( x = 4 \).

### Step 4: Test intervals

We will test the intervals \( (-\infty, -4) \), \( (-4, 4) \), and \( (4, \infty) \):

1. For \( x < -4 \) (e.g., \( x = -5 \)):
\((x - 4)(x + 4) > 0\) (positive)

2. For \( -4 < x < 4 \) (e.g., \( x = 0 \)):
\((x - 4)(x + 4) < 0\) (negative)

3. For \( x > 4 \) (e.g., \( x = 5 \)):
\((x - 4)(x + 4) > 0\) (positive)

Thus, the solution to \( (x - 4)(x + 4) < 0 \) is:

\[ -4 < x < 4 \]

### Step 5: Combine the results

From Step 1, we have \( x > 0 \). Combining this with the result from Step 4:

\[ 0 < x < 4 \]

### Step 6: Identify integer solutions

The integer solutions in the interval \( (0, 4) \) are:

\[ x = 1, 2, 3 \]

Thus, there are **3 integer solutions**.

### Final Answer

The number of integer solutions is **3**.

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