To solve the problem, we need to find the roots of the given quadratic equations and match them with their corresponding solution sets.
Let's begin by solving each quadratic equation step-by-step:
1. Solve [tex]\( a^2 - 9a + 14 = 0 \)[/tex]
This is in the standard form [tex]\( ax^2 + bx + c = 0 \)[/tex].
The solutions [tex]\( a \)[/tex] are given by the quadratic formula:
[tex]\[
a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
For [tex]\( a^2 - 9a + 14 = 0 \)[/tex]:
[tex]\[
a = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(14)}}{2(1)}
\][/tex]
[tex]\[
a = \frac{9 \pm \sqrt{81 - 56}}{2}
\][/tex]
[tex]\[
a = \frac{9 \pm \sqrt{25}}{2}
\][/tex]
[tex]\[
a = \frac{9 \pm 5}{2}
\][/tex]
So, the solutions are:
[tex]\[
a = \frac{9 + 5}{2} = 7 \quad \text{and} \quad a = \frac{9 - 5}{2} = 2
\][/tex]
Therefore, the solution set for [tex]\( a^2 - 9a + 14 = 0 \)[/tex] is [tex]\( \{7, 2\} \)[/tex].
2. Solve [tex]\( a^2 - 5a - 14 = 0 \)[/tex]
Again using the quadratic formula:
[tex]\[
a = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(-14)}}{2(1)}
\][/tex]
[tex]\[
a = \frac{5 \pm \sqrt{25 + 56}}{2}
\][/tex]
[tex]\[
a = \frac{5 \pm \sqrt{81}}{2}
\][/tex]
[tex]\[
a = \frac{5 \pm 9}{2}
\][/tex]
So, the solutions are:
[tex]\[
a = \frac{5 + 9}{2} = 7 \quad \text{and} \quad a = \frac{5 - 9}{2} = -2
\][/tex]
Therefore, the solution set for [tex]\( a^2 - 5a - 14 = 0 \)[/tex] is [tex]\( \{7, -2\} \)[/tex].
Now, let's match these solution sets with the given equations.
1. [tex]\( a^2 - 9a + 14 = \{2, 7\} \)[/tex]
2. [tex]\( a^2 - 5a - 14 = \{7, -2\} \)[/tex]
So, the final matching is:
- [tex]\( a^2 - 9a + 14 = 0 \)[/tex] matches with [tex]\( \{2, 7\} \)[/tex]
- [tex]\( a^2 - 5a - 14 = 0 \)[/tex] matches with [tex]\( \{7, -2\} \)[/tex]
