Question

Each of the following quadratic equations can be solved by factoring. Which equation has only one distinct solution?(1 point)
Responses

x2+4x+4=0
x
2
+
4
x
+
4
=
0
x squared plus 4 x plus 4 equals 0

x2−5x+4=0
x
2
−
5
x
+
4
=
0
x squared minus 5 x plus 4 equals 0

x2−4=0
x
2
−
4
=
0
x squared minus 4 equals 0

x2+5x+4=0

Answer 1

To determine which quadratic equation has only one distinct solution, we need to check for equations that can be factored into the form \((x - r)^2 = 0\), where \(r\) is a repeated root.

1. **\(x^2 + 4x + 4 = 0\)**
This can be factored as \((x + 2)(x + 2) = (x + 2)^2 = 0\).
**Distinct solution**: \(x = -2\) (one solution).

2. **\(x^2 - 5x + 4 = 0\)**
This can be factored as \((x - 1)(x - 4) = 0\).
**Distinct solutions**: \(x = 1\) and \(x = 4\) (two solutions).

3. **\(x^2 - 4 = 0\)**
This can be factored as \((x - 2)(x + 2) = 0\).
**Distinct solutions**: \(x = 2\) and \(x = -2\) (two solutions).

4. **\(x^2 + 5x + 4 = 0\)**
This can be factored as \((x + 1)(x + 4) = 0\).
**Distinct solutions**: \(x = -1\) and \(x = -4\) (two solutions).

The equation that has only one distinct solution is:
**\(x^2 + 4x + 4 = 0\)**.