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.

1 answer

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.