High School

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------------------------------------------------ Solve the equation using the quadratic formula:

[tex]15x^2 + 13x = 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 :

Sure! Let's solve the quadratic equation [tex]\(15x^2 + 13x = 0\)[/tex] using the quadratic formula.

The quadratic formula is given by:

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

For the equation [tex]\(15x^2 + 13x = 0\)[/tex], the coefficients are:
- [tex]\(a = 15\)[/tex]
- [tex]\(b = 13\)[/tex]
- [tex]\(c = 0\)[/tex]

Step 1: Calculate the Discriminant

The discriminant [tex]\((b^2 - 4ac)\)[/tex] is essential to use the quadratic formula:

[tex]\[
b^2 - 4ac = 13^2 - 4 \times 15 \times 0
\][/tex]
[tex]\[
b^2 - 4ac = 169 - 0 = 169
\][/tex]

Step 2: Use the Quadratic Formula

We plug the values into the quadratic formula:

[tex]\[
x = \frac{-13 \pm \sqrt{169}}{2 \times 15}
\][/tex]

Since [tex]\(\sqrt{169} = 13\)[/tex], we continue:

[tex]\[
x = \frac{-13 \pm 13}{30}
\][/tex]

Step 3: Calculate the Two Solutions

1. For [tex]\(x_1\)[/tex]:

[tex]\[
x_1 = \frac{-13 + 13}{30} = \frac{0}{30} = 0
\][/tex]

2. For [tex]\(x_2\)[/tex]:

[tex]\[
x_2 = \frac{-13 - 13}{30} = \frac{-26}{30} = -\frac{13}{15}
\][/tex]

Conclusion:

The solutions to the equation [tex]\(15x^2 + 13x = 0\)[/tex] are [tex]\(x = 0\)[/tex] and [tex]\(x = -\frac{13}{15}\)[/tex].

Thus, the best choice from the given options is A: [tex]\(x = -\frac{13}{15}, 0\)[/tex].