The curvey `y = (a +1) x ^(2) + 2` meets the curve `y = ax +3, a ne -1` in exactly one point, then `a^(2) =`

Answer

To solve the problem, we need to find the value of \( a^2 \) given that the curves \( y = (a + 1)x^2 + 2 \) and \( y = ax + 3 \) meet at exactly one point.

### Step-by-Step Solution:

1. **Set the equations equal to each other**:
Since the two curves meet at a point, we can set their equations equal to each other:
\[ (a + 1)x^2 + 2 = ax + 3 \]

2. **Rearrange the equation**:
Rearranging gives us:
\[ (a + 1)x^2 - ax + 2 - 3 = 0 \]
Simplifying this, we get:
\[ (a + 1)x^2 - ax - 1 = 0 \]

3. **Identify the coefficients**:
The equation is now in the standard quadratic form \( Ax^2 + Bx + C = 0 \), where:
- \( A = a + 1 \)
- \( B = -a \)
- \( C = -1 \)

4. **Condition for exactly one solution**:
For the quadratic equation to have exactly one solution, the discriminant must be zero. The discriminant \( D \) is given by:
\[ D = B^2 - 4AC \]
Substituting the coefficients, we have:
\[ D = (-a)^2 - 4(a + 1)(-1) \]
Simplifying this gives:
\[ D = a^2 + 4(a + 1) = a^2 + 4a + 4 \]

5. **Set the discriminant to zero**:
For there to be exactly one solution:
\[ a^2 + 4a + 4 = 0 \]

6. **Factor the quadratic equation**:
This can be factored as:
\[ (a + 2)^2 = 0 \]

7. **Solve for \( a \)**:
Taking the square root of both sides gives:
\[ a + 2 = 0 \implies a = -2 \]

8. **Find \( a^2 \)**:
Now we calculate \( a^2 \):
\[ a^2 = (-2)^2 = 4 \]

### Final Answer:
\[ a^2 = 4 \]