Question

Solve the​ equation, and check the solutions.
18p squaredequals21minus33p
Question content area bottom
Part 1
The solution set is StartSet {Insert answer here} EndSet
  
.
​(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:

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