Question
Complete the work shown to answer the question.
p2 – 14p – 72 = 0
p2 – 14p = 72
p2 – 14p + 49 = 72 + 49
Which describes the solutions of the equation?
Since 7 and 11 are both rational, the sum and difference are rational.
Since 14 and 11 are both rational, the sum and difference are rational.
Since 7 is rational and StartRoot 11 EndRoot is irrational, the sum and difference are irrational.
Since StartRoot 7 EndRoot is irrational and 11 is rational, the sum and difference are irrational.
p2 – 14p – 72 = 0
p2 – 14p = 72
p2 – 14p + 49 = 72 + 49
Which describes the solutions of the equation?
Since 7 and 11 are both rational, the sum and difference are rational.
Since 14 and 11 are both rational, the sum and difference are rational.
Since 7 is rational and StartRoot 11 EndRoot is irrational, the sum and difference are irrational.
Since StartRoot 7 EndRoot is irrational and 11 is rational, the sum and difference are irrational.
Answers
GPT-4o mini
Let's first complete the work shown in the equation given:
1. Start with the original equation:
\[
p^2 - 14p - 72 = 0
\]
2. Move 72 to the right side:
\[
p^2 - 14p = 72
\]
3. To complete the square, add \(\left(\frac{14}{2}\right)^2 = 49\) to both sides:
\[
p^2 - 14p + 49 = 72 + 49
\]
4. This simplifies to:
\[
(p - 7)^2 = 121
\]
5. Now, take the square root of both sides:
\[
p - 7 = \pm 11
\]
6. This leads to two possible solutions:
\[
p - 7 = 11 \quad \Rightarrow \quad p = 18
\]
\[
p - 7 = -11 \quad \Rightarrow \quad p = -4
\]
Now, regarding the descriptions of the solutions:
The solutions we found are \(p = 18\) and \(p = -4\), both of which are rational numbers.
1. Options:
- Since 7 and 11 are both rational, the sum and difference are rational.
- Since 14 and 11 are both rational, the sum and difference are rational.
- Since 7 is rational and \(\sqrt{11}\) is irrational, the sum and difference are irrational.
- Since \(\sqrt{7}\) is irrational and 11 is rational, the sum and difference are irrational.
Evaluating these options, we find that the first two options apply since both 7 and 11 (as well as 14) are rational numbers. The sum (like \(7 + 11\)) and the difference (like \(7 - 11\)) of rational numbers is always rational.
Thus, both the following answers are correct:
- **Since 7 and 11 are both rational, the sum and difference are rational.**
- **Since 14 and 11 are both rational, the sum and difference are rational.**
Depending on the context and the required choice, you may select either option.
1. Start with the original equation:
\[
p^2 - 14p - 72 = 0
\]
2. Move 72 to the right side:
\[
p^2 - 14p = 72
\]
3. To complete the square, add \(\left(\frac{14}{2}\right)^2 = 49\) to both sides:
\[
p^2 - 14p + 49 = 72 + 49
\]
4. This simplifies to:
\[
(p - 7)^2 = 121
\]
5. Now, take the square root of both sides:
\[
p - 7 = \pm 11
\]
6. This leads to two possible solutions:
\[
p - 7 = 11 \quad \Rightarrow \quad p = 18
\]
\[
p - 7 = -11 \quad \Rightarrow \quad p = -4
\]
Now, regarding the descriptions of the solutions:
The solutions we found are \(p = 18\) and \(p = -4\), both of which are rational numbers.
1. Options:
- Since 7 and 11 are both rational, the sum and difference are rational.
- Since 14 and 11 are both rational, the sum and difference are rational.
- Since 7 is rational and \(\sqrt{11}\) is irrational, the sum and difference are irrational.
- Since \(\sqrt{7}\) is irrational and 11 is rational, the sum and difference are irrational.
Evaluating these options, we find that the first two options apply since both 7 and 11 (as well as 14) are rational numbers. The sum (like \(7 + 11\)) and the difference (like \(7 - 11\)) of rational numbers is always rational.
Thus, both the following answers are correct:
- **Since 7 and 11 are both rational, the sum and difference are rational.**
- **Since 14 and 11 are both rational, the sum and difference are rational.**
Depending on the context and the required choice, you may select either option.