Answer :
Let's solve each quadratic equation step-by-step and pair them with their corresponding solution sets derived earlier:
1. Equation: [tex]\(a^2 - 9a + 14 = 0\)[/tex]
To solve this quadratic equation, we need to factor it:
[tex]\[ a^2 - 9a + 14 = (a - 2)(a - 7) = 0 \][/tex]
Setting each factor equal to zero gives us the solutions:
[tex]\[ a - 2 = 0 \quad \Rightarrow \quad a = 2 \][/tex]
[tex]\[ a - 7 = 0 \quad \Rightarrow \quad a = 7 \][/tex]
Therefore, the solution set is [tex]\([2, 7]\)[/tex].
2. Equation: [tex]\(a^2 + 9a + 14 = 0\)[/tex]
To solve this quadratic equation, we factor it:
[tex]\[ a^2 + 9a + 14 = (a + 2)(a + 7) = 0 \][/tex]
Setting each factor equal to zero gives us the solutions:
[tex]\[ a + 2 = 0 \quad \Rightarrow \quad a = -2 \][/tex]
[tex]\[ a + 7 = 0 \quad \Rightarrow \quad a = -7 \][/tex]
Therefore, the solution set is [tex]\([-7, -2]\)[/tex].
3. Equation: [tex]\(a^2 + 3a - 10 = 0\)[/tex]
To solve this quadratic equation, we factor it:
[tex]\[ a^2 + 3a - 10 = (a + 5)(a - 2) = 0 \][/tex]
Setting each factor equal to zero gives us the solutions:
[tex]\[ a + 5 = 0 \quad \Rightarrow \quad a = -5 \][/tex]
[tex]\[ a - 2 = 0 \quad \Rightarrow \quad a = 2 \][/tex]
Therefore, the solution set is [tex]\([-5, 2]\)[/tex].
4. Equation: [tex]\(a^2 + 5a - 14 = 0\)[/tex]
To solve this quadratic equation, we factor it:
[tex]\[ a^2 + 5a - 14 = (a + 7)(a - 2) = 0 \][/tex]
Setting each factor equal to zero gives us the solutions:
[tex]\[ a + 7 = 0 \quad \Rightarrow \quad a = -7 \][/tex]
[tex]\[ a - 2 = 0 \quad \Rightarrow \quad a = 2 \][/tex]
Therefore, the solution set is [tex]\([-7, 2]\)[/tex].
5. Equation: [tex]\(a^2 - 5a - 14 = 0\)[/tex]
To solve this quadratic equation, we factor it:
[tex]\[ a^2 - 5a - 14 = (a - 7)(a + 2) = 0 \][/tex]
Setting each factor equal to zero gives us the solutions:
[tex]\[ a - 7 = 0 \quad \Rightarrow \quad a = 7 \][/tex]
[tex]\[ a + 2 = 0 \quad \Rightarrow \quad a = -2 \][/tex]
Therefore, the solution set is [tex]\([-2, 7]\)[/tex].
Now we can match each equation with its solution set:
- [tex]\(a^2 - 9a + 14 = 0\)[/tex] has the solution set [tex]\([2, 7]\)[/tex].
- [tex]\(a^2 + 9a + 14 = 0\)[/tex] has the solution set [tex]\([-7, -2]\)[/tex].
- [tex]\(a^2 + 3a - 10 = 0\)[/tex] has the solution set [tex]\([-5, 2]\)[/tex].
- [tex]\(a^2 + 5a - 14 = 0\)[/tex] has the solution set [tex]\([-7, 2]\)[/tex].
- [tex]\(a^2 - 5a - 14 = 0\)[/tex] has the solution set [tex]\([-2, 7]\)[/tex].
1. Equation: [tex]\(a^2 - 9a + 14 = 0\)[/tex]
To solve this quadratic equation, we need to factor it:
[tex]\[ a^2 - 9a + 14 = (a - 2)(a - 7) = 0 \][/tex]
Setting each factor equal to zero gives us the solutions:
[tex]\[ a - 2 = 0 \quad \Rightarrow \quad a = 2 \][/tex]
[tex]\[ a - 7 = 0 \quad \Rightarrow \quad a = 7 \][/tex]
Therefore, the solution set is [tex]\([2, 7]\)[/tex].
2. Equation: [tex]\(a^2 + 9a + 14 = 0\)[/tex]
To solve this quadratic equation, we factor it:
[tex]\[ a^2 + 9a + 14 = (a + 2)(a + 7) = 0 \][/tex]
Setting each factor equal to zero gives us the solutions:
[tex]\[ a + 2 = 0 \quad \Rightarrow \quad a = -2 \][/tex]
[tex]\[ a + 7 = 0 \quad \Rightarrow \quad a = -7 \][/tex]
Therefore, the solution set is [tex]\([-7, -2]\)[/tex].
3. Equation: [tex]\(a^2 + 3a - 10 = 0\)[/tex]
To solve this quadratic equation, we factor it:
[tex]\[ a^2 + 3a - 10 = (a + 5)(a - 2) = 0 \][/tex]
Setting each factor equal to zero gives us the solutions:
[tex]\[ a + 5 = 0 \quad \Rightarrow \quad a = -5 \][/tex]
[tex]\[ a - 2 = 0 \quad \Rightarrow \quad a = 2 \][/tex]
Therefore, the solution set is [tex]\([-5, 2]\)[/tex].
4. Equation: [tex]\(a^2 + 5a - 14 = 0\)[/tex]
To solve this quadratic equation, we factor it:
[tex]\[ a^2 + 5a - 14 = (a + 7)(a - 2) = 0 \][/tex]
Setting each factor equal to zero gives us the solutions:
[tex]\[ a + 7 = 0 \quad \Rightarrow \quad a = -7 \][/tex]
[tex]\[ a - 2 = 0 \quad \Rightarrow \quad a = 2 \][/tex]
Therefore, the solution set is [tex]\([-7, 2]\)[/tex].
5. Equation: [tex]\(a^2 - 5a - 14 = 0\)[/tex]
To solve this quadratic equation, we factor it:
[tex]\[ a^2 - 5a - 14 = (a - 7)(a + 2) = 0 \][/tex]
Setting each factor equal to zero gives us the solutions:
[tex]\[ a - 7 = 0 \quad \Rightarrow \quad a = 7 \][/tex]
[tex]\[ a + 2 = 0 \quad \Rightarrow \quad a = -2 \][/tex]
Therefore, the solution set is [tex]\([-2, 7]\)[/tex].
Now we can match each equation with its solution set:
- [tex]\(a^2 - 9a + 14 = 0\)[/tex] has the solution set [tex]\([2, 7]\)[/tex].
- [tex]\(a^2 + 9a + 14 = 0\)[/tex] has the solution set [tex]\([-7, -2]\)[/tex].
- [tex]\(a^2 + 3a - 10 = 0\)[/tex] has the solution set [tex]\([-5, 2]\)[/tex].
- [tex]\(a^2 + 5a - 14 = 0\)[/tex] has the solution set [tex]\([-7, 2]\)[/tex].
- [tex]\(a^2 - 5a - 14 = 0\)[/tex] has the solution set [tex]\([-2, 7]\)[/tex].
