To determine which equation has only one distinct solution, we need to find the discriminant (D) of each quadratic equation. The discriminant can be calculated using the formula:
\[ D = b^2 - 4ac \]
For a quadratic equation of the form \( ax^2 + bx + c = 0 \):
- If \( D > 0 \): There are two distinct real solutions.
- If \( D = 0 \): There is one distinct real solution (this is what we're looking for).
- If \( D < 0 \): There are no real solutions.
Let's analyze each equation:
-
\( x^2 - 4 = 0 \)
- Here, \( a = 1, b = 0, c = -4 \).
- Discriminant: \( D = 0^2 - 4(1)(-4) = 0 + 16 = 16 \) (two solutions)
-
\( x^2 + 4x + 4 = 0 \)
- Here, \( a = 1, b = 4, c = 4 \).
- Discriminant: \( D = 4^2 - 4(1)(4) = 16 - 16 = 0 \) (one solution)
-
\( x^2 - 5x + 4 = 0 \)
- Here, \( a = 1, b = -5, c = 4 \).
- Discriminant: \( D = (-5)^2 - 4(1)(4) = 25 - 16 = 9 \) (two solutions)
-
\( x^2 + 5x + 4 = 0 \)
- Here, \( a = 1, b = 5, c = 4 \).
- Discriminant: \( D = 5^2 - 4(1)(4) = 25 - 16 = 9 \) (two solutions)
Now, based on the discriminants calculated:
- The equation \( x^2 + 4x + 4 = 0 \) is the only one that has one distinct solution (since its discriminant is 0).
3 answers