High School

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 :

To solve the quadratic equation [tex]\( 15x^2 + 13x = 0 \)[/tex] using the quadratic formula, follow these steps:

1. Write down the quadratic formula:

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

2. Identify the coefficients [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] from the equation [tex]\( 15x^2 + 13x + 0 = 0 \)[/tex]:

[tex]\[
a = 15, \quad b = 13, \quad c = 0
\][/tex]

3. Calculate the discriminant [tex]\( \Delta \)[/tex]:

[tex]\[
\Delta = b^2 - 4ac = 13^2 - (4 \cdot 15 \cdot 0) = 169
\][/tex]

4. Find the square root of the discriminant:

[tex]\[
\sqrt{169} = 13
\][/tex]

5. Substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( \sqrt{\Delta} \)[/tex] into the quadratic formula to find the roots:

[tex]\[
x = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-13 \pm 13}{2 \cdot 15}
\][/tex]

6. Solve for the two values of [tex]\( x \)[/tex]:

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

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

Therefore, the solutions to the equation [tex]\( 15x^2 + 13x = 0 \)[/tex] are:

[tex]\[
x = 0 \quad \text{and} \quad x = -\frac{13}{15}
\][/tex]

So, the correct answer from the choices provided is:

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

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