Solve the​ equation, and check the solutions.

Question

18p squaredequals21minus33p
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Part 1
The solution set is StartSet {Insert answer here} EndSet
  
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​(Use a comma to separate answers as needed. Type an integer or a simplified​ fraction.)

Answer 1

To solve the equation \( 18p^2 = 21 - 33p \), we can start by rearranging it into standard quadratic form:

1. Move all terms to one side of the equation:

   \[ 18p^2 + 33p - 21 = 0 \]

Now, we will solve this quadratic equation using the quadratic formula:

\[ p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

where \( a = 18 \), \( b = 33 \), and \( c = -21 \).

2. Calculate the discriminant \( b^2 - 4ac \):

   \[ b^2 = 33^2 = 1089 \] \[ 4ac = 4 \cdot 18 \cdot (-21) = -1512 \] \[ b^2 - 4ac = 1089 + 1512 = 2601 \]

3. Now plug the values into the quadratic formula:

   \[ p = \frac{-33 \pm \sqrt{2601}}{2 \cdot 18} \]

4. Simplifying \( \sqrt{2601} \):

   \[ \sqrt{2601} = 51 \quad (\text{because } 51^2 = 2601) \]

5. Plugging this back in:

   \[ p = \frac{-33 \pm 51}{36} \]

This gives us two potential solutions:

• For the \( + \):

  \[ p = \frac{-33 + 51}{36} = \frac{18}{36} = \frac{1}{2} \]

• For the \( - \):

  \[ p = \frac{-33 - 51}{36} = \frac{-84}{36} = -\frac{7}{3} \]

Thus, the solution set is:

\[ { \frac{1}{2}, -\frac{7}{3} } \]

Checking the Solutions

For \( p = \frac{1}{2} \):

\[ 18p^2 = 18 \left( \frac{1}{2} \right)^2 = 18 \cdot \frac{1}{4} = \frac{18}{4} = \frac{9}{2} \] \[ 21 - 33p = 21 - 33 \cdot \frac{1}{2} = 21 - \frac{33}{2} = \frac{42}{2} - \frac{33}{2} = \frac{9}{2} \]

Both sides are equal, so \( p = \frac{1}{2} \) is correct.

For \( p = -\frac{7}{3} \):

\[ 18p^2 = 18 \left( -\frac{7}{3} \right)^2 = 18 \cdot \frac{49}{9} = 2 \cdot 49 = 98 \] \[ 21 - 33p = 21 - 33 \left( -\frac{7}{3} \right) = 21 + 77 = 98 \]

Both sides are equal, so \( p = -\frac{7}{3} \) is also correct.

Final Solution Set

The solution set is:

\[ { \frac{1}{2}, -\frac{7}{3} } \]