Which of the following equations has no solution for a?

Options

A. `a^2 - 6a + 7 = 0`
B. `a^2 + 6a - 7 = 0`
C. `a^2 + 4a + 3 = 0`
D. `a^2 - 4a + 5 = 0`

Answer

To determine which of the given equations has no solution for \( a \), we will analyze each equation using the discriminant method. The discriminant \( D \) of a quadratic equation of the form \( Ax^2 + Bx + C = 0 \) is given by the formula:

\[ D = B^2 - 4AC \]

If \( D < 0 \), the quadratic equation has no real solutions. If \( D \geq 0 \), the quadratic equation has real solutions.

Let's evaluate each option:

### Option 1: \( a^2 - 6a + 7 = 0 \)

1. Identify \( A = 1 \), \( B = -6 \), \( C = 7 \).
2. Calculate the discriminant:
\[ D = (-6)^2 - 4 \cdot 1 \cdot 7 = 36 - 28 = 8 \]
3. Since \( D = 8 \) (which is greater than 0), this equation has real solutions.

### Option 2: \( a^2 + 6a - 7 = 0 \)

1. Identify \( A = 1 \), \( B = 6 \), \( C = -7 \).
2. Calculate the discriminant:
\[ D = 6^2 - 4 \cdot 1 \cdot (-7) = 36 + 28 = 64 \]
3. Since \( D = 64 \) (which is greater than 0), this equation has real solutions.

### Option 3: \( a^2 + 4a + 3 = 0 \)

1. Identify \( A = 1 \), \( B = 4 \), \( C = 3 \).
2. Calculate the discriminant:
\[ D = 4^2 - 4 \cdot 1 \cdot 3 = 16 - 12 = 4 \]
3. Since \( D = 4 \) (which is greater than 0), this equation has real solutions.

### Option 4: \( a^2 - 4a + 5 = 0 \)

1. Identify \( A = 1 \), \( B = -4 \), \( C = 5 \).
2. Calculate the discriminant:
\[ D = (-4)^2 - 4 \cdot 1 \cdot 5 = 16 - 20 = -4 \]
3. Since \( D = -4 \) (which is less than 0), this equation has no real solutions.

### Conclusion

The equation that has no solution for \( a \) is:

**Option 4: \( a^2 - 4a + 5 = 0 \)**

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