Let's go through each equation one by one to determine if it has only one solution.
1. [tex]\( x^2 - x - 6 = 0 \)[/tex]:
This equation can be factored as:
[tex]\[
(x - 3)(x + 2) = 0
\][/tex]
Setting each factor to zero gives the solutions:
[tex]\[
x = 3 \quad \text{or} \quad x = -2
\][/tex]
Thus, this equation has two solutions.
2. [tex]\( 5x^2 + 20x + 20 = 0 \)[/tex]:
This equation can be simplified by dividing through by 5:
[tex]\[
x^2 + 4x + 4 = 0
\][/tex]
Notice that this can be written as:
[tex]\[
(x + 2)^2 = 0
\][/tex]
Setting the factor to zero gives the solution:
[tex]\[
x = -2
\][/tex]
Thus, this equation has one solution.
3. [tex]\( 9x^2 - 25 = 0 \)[/tex]:
This is a difference of squares:
[tex]\[
(3x + 5)(3x - 5) = 0
\][/tex]
Setting each factor to zero gives the solutions:
[tex]\[
x = \frac{5}{3} \quad \text{or} \quad x = -\frac{5}{3}
\][/tex]
Thus, this equation has two solutions.
4. [tex]\( 4x^2 + 4x = 0 \)[/tex]:
This equation can be factored by factoring out a common term:
[tex]\[
4x(x + 1) = 0
\][/tex]
Setting each factor to zero gives the solutions:
[tex]\[
x = 0 \quad \text{or} \quad x = -1
\][/tex]
Thus, this equation has two solutions.
5. [tex]\( x^2 + 6x + 9 = 0 \)[/tex]:
Notice that this can be written as:
[tex]\[
(x + 3)^2 = 0
\][/tex]
Setting the factor to zero gives the solution:
[tex]\[
x = -3
\][/tex]
Thus, this equation has one solution.
Therefore, the equations with only one solution are:
[tex]\[
\boxed{2 \text{ and } 5}
\][/tex]
