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Solve the equation by using the quadratic formula:

\[ 15x^2 + 13x = 0 \]

a. \( x = -\frac{13}{15}, 0 \)
b. \( x = 0 \)
c. \( x = \frac{13}{15}, 0 \)
d. \( x = \pm \frac{13}{15} \)

Answer :

The correct answer is: [tex]x = -13/15, x = 0.[/tex]

To solve the quadratic equation [tex]15x^2 + 13x = 0[/tex]

Using the quadratic formula x = (-b ± [tex]\sqrt{ (b^2 - 4ac}[/tex])) / [tex]2a[/tex], we first identify the values of a, b, and c. In this equation, a = 15, b = 13, and c = 0.

Substituting these values into the quadratic formula, we get:

x = ([tex]-13[/tex] ± [tex]\sqrt{(13^2 - 4(15)(0)))} / 2(15)[/tex]

x = ([tex]-13[/tex] ± [tex]\sqrt{(169)) / 30}[/tex]

x = [tex](-13[/tex] ± [tex]13) / 30[/tex]

Simplifying this expression, we find that there are two possible solutions for x:

[tex]x = -13/15[/tex] or [tex]x = 0[/tex]

Therefore, the correct answer is:

x = -13/15, x = 0.

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