The set of all real numbers x such that `x^(2)+2x,2x+3` and `x^(2)+3x+8` are the sides of a triangle is …………

Answer

To determine the set of all real numbers \( x \) such that \( x^2 + 2x \), \( 2x + 3 \), and \( x^2 + 3x + 8 \) can be the sides of a triangle, we need to apply the triangle inequality. The triangle inequality states that for any triangle with sides \( a \), \( b \), and \( c \), the following conditions must hold:

1. \( a + b > c \)
2. \( a + c > b \)
3. \( b + c > a \)

In our case, we will let:
- \( a = x^2 + 2x \)
- \( b = 2x + 3 \)
- \( c = x^2 + 3x + 8 \)

### Step 1: Apply the first triangle inequality \( a + b > c \)

\[ (x^2 + 2x) + (2x + 3) > (x^2 + 3x + 8) \]

Simplifying this:

\[ x^2 + 2x + 2x + 3 > x^2 + 3x + 8 \]

This simplifies to:

\[ 4x + 3 > 3x + 8 \]

Subtract \( 3x \) from both sides:

\[ x + 3 > 8 \]

Subtract 3 from both sides:

\[ x > 5 \]

### Step 2: Apply the second triangle inequality \( a + c > b \)

\[ (x^2 + 2x) + (x^2 + 3x + 8) > (2x + 3) \]

Simplifying this:

\[ x^2 + 2x + x^2 + 3x + 8 > 2x + 3 \]

This simplifies to:

\[ 2x^2 + 5x + 8 > 2x + 3 \]

Subtract \( 2x + 3 \) from both sides:

\[ 2x^2 + 3x + 5 > 0 \]

This is a quadratic inequality. The discriminant \( D \) of \( 2x^2 + 3x + 5 \) is:

\[ D = b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot 5 = 9 - 40 = -31 \]

Since the discriminant is negative, the quadratic \( 2x^2 + 3x + 5 \) is always positive for all real \( x \).

### Step 3: Apply the third triangle inequality \( b + c > a \)

\[ (2x + 3) + (x^2 + 3x + 8) > (x^2 + 2x) \]

Simplifying this:

\[ 2x + 3 + x^2 + 3x + 8 > x^2 + 2x \]

This simplifies to:

\[ 5x + 11 > 2x \]

Subtract \( 2x \) from both sides:

\[ 3x + 11 > 0 \]

This simplifies to:

\[ x > -\frac{11}{3} \]

### Final Step: Combine the inequalities

From the three inequalities we have:

1. \( x > 5 \)
2. \( 2x^2 + 3x + 5 > 0 \) (always true)
3. \( x > -\frac{11}{3} \)

The most restrictive condition is \( x > 5 \).

Thus, the set of all real numbers \( x \) such that \( x^2 + 2x \), \( 2x + 3 \), and \( x^2 + 3x + 8 \) can be the sides of a triangle is:

\[ \boxed{(5, \infty)} \]