Solve the equation using the quadratic formula:

[tex]15\pi^2 + 13\pi = 0[/tex]

a. [tex]x = -\frac{13}{15}, 0[/tex]
b. [tex]x = 0[/tex]
c. [tex]x = \frac{13}{15}, 0[/tex]
d. [tex]x = \pm \frac{13}{15}[/tex]

Please select the best answer from the choices provided:
A
B
C
D

Answer :

To solve the equation [tex]\(15\pi^2 + 13\pi = 0\)[/tex], we can follow these steps:

1. Identify it as a quadratic equation:
The equation is already in a quadratic form with [tex]\(\pi\)[/tex] as the variable. We can rewrite it as:
[tex]\[
15\pi^2 + 13\pi = 0
\][/tex]
Here, [tex]\(a = 15\)[/tex], [tex]\(b = 13\)[/tex], and [tex]\(c = 0\)[/tex].

2. Factor the equation:
Notice that both terms in the equation have a common factor, [tex]\(\pi\)[/tex]. We can factor [tex]\(\pi\)[/tex] out:
[tex]\[
\pi(15\pi + 13) = 0
\][/tex]

3. Set each factor to zero:
To find the solutions, set each factor equal to zero:
- First factor: [tex]\(\pi = 0\)[/tex]
- Second factor: [tex]\(15\pi + 13 = 0\)[/tex]

4. Solve each equation:
- For [tex]\(\pi = 0\)[/tex], we simply have [tex]\(\pi = 0\)[/tex].
- For [tex]\(15\pi + 13 = 0\)[/tex], solve for [tex]\(\pi\)[/tex]:
[tex]\[
15\pi = -13
\][/tex]
[tex]\[
\pi = -\frac{13}{15}
\][/tex]

5. Conclude the solutions:
The solutions to the equation are [tex]\(\pi = 0\)[/tex] and [tex]\(\pi = -\frac{13}{15}\)[/tex].

Now, let's match these solutions to the choices provided:

a. [tex]\(x = -\frac{13}{15}, 0\)[/tex]

b. [tex]\(x = 0\)[/tex]

c. [tex]\(x = \frac{13}{15}, 0\)[/tex]

d. [tex]\(x = \pm \frac{13}{15}\)[/tex]

The correct answer is option a: [tex]\(x = -\frac{13}{15}, 0\)[/tex].