Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.
Sagot :
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]
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]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.
