The numebr of solution (s) of the inequation
`sqrt(3x^(2)+6x+7)+sqrt(5x^(2)+10x+14)le4-2x-x^(2)`, is

Options

A. 1
B. 2
C. 4
D. infinitely many

Answer

To solve the inequality

\[ \sqrt{3x^2 + 6x + 7} + \sqrt{5x^2 + 10x + 14} \leq 4 - 2x - x^2, \]

we will analyze both sides step by step.

### Step 1: Analyze the Left-Hand Side (LHS)

The first term in the LHS is

\[ \sqrt{3x^2 + 6x + 7}. \]

We can rewrite the expression inside the square root:

\[ 3x^2 + 6x + 7 = 3(x^2 + 2x) + 7 = 3(x^2 + 2x + 1 - 1) + 7 = 3((x + 1)^2 - 1) + 7 = 3(x + 1)^2 + 4. \]

Thus, we have:

\[ \sqrt{3x^2 + 6x + 7} = \sqrt{3(x + 1)^2 + 4}. \]

This expression is always greater than or equal to 2 because the minimum value occurs when \(x = -1\), giving us:

\[ \sqrt{3(0) + 4} = \sqrt{4} = 2. \]

### Step 2: Analyze the Second Term in LHS

Now, consider the second term:

\[ \sqrt{5x^2 + 10x + 14}. \]

We can factor this similarly:

\[ 5x^2 + 10x + 14 = 5(x^2 + 2x) + 14 = 5((x + 1)^2 - 1) + 14 = 5(x + 1)^2 + 9. \]

Thus, we have:

\[ \sqrt{5x^2 + 10x + 14} = \sqrt{5(x + 1)^2 + 9}. \]

This expression is always greater than or equal to 3 because the minimum value occurs when \(x = -1\), giving us:

\[ \sqrt{5(0) + 9} = \sqrt{9} = 3. \]

### Step 3: Combine the LHS

Now, combining both terms in the LHS:

\[ \sqrt{3x^2 + 6x + 7} + \sqrt{5x^2 + 10x + 14} \geq 2 + 3 = 5. \]

### Step 4: Analyze the Right-Hand Side (RHS)

Now let's analyze the RHS:

\[ 4 - 2x - x^2. \]

This is a downward-facing quadratic equation. To find its maximum value, we can complete the square:

\[ 4 - 2x - x^2 = -(x^2 + 2x - 4) = -((x + 1)^2 - 5) = 5 - (x + 1)^2. \]

The maximum value occurs when \((x + 1)^2 = 0\), which gives:

\[ 5 - 0 = 5. \]

### Step 5: Setting LHS and RHS

Now, we have established:

- LHS \( \geq 5 \)
- RHS \( \leq 5 \)

The only point where they can be equal is when both are equal to 5. This occurs when \(x = -1\).

### Conclusion

Thus, the only solution to the inequality is:

\[ \text{The number of solutions is } 1. \]