Answer :
To solve the equation [tex]\(15x^2 + 13x = 0\)[/tex] using the quadratic formula, we need to identify the coefficients in the standard form [tex]\(ax^2 + bx + c = 0\)[/tex]. For this equation:
- [tex]\(a = 15\)[/tex]
- [tex]\(b = 13\)[/tex]
- [tex]\(c = 0\)[/tex]
The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Let's calculate the solutions step-by-step:
1. Calculate the Discriminant:
[tex]\[
b^2 - 4ac = 13^2 - 4 \times 15 \times 0 = 169 - 0 = 169
\][/tex]
2. Solve for [tex]\(x\)[/tex] using the quadratic formula:
[tex]\[
x = \frac{-13 \pm \sqrt{169}}{2 \times 15}
\][/tex]
- Since [tex]\(\sqrt{169} = 13\)[/tex], we compute the two possible values for [tex]\(x\)[/tex]:
- First solution:
[tex]\[
x_1 = \frac{-13 + 13}{30} = \frac{0}{30} = 0
\][/tex]
- Second solution:
[tex]\[
x_2 = \frac{-13 - 13}{30} = \frac{-26}{30} = -\frac{26}{30} = -\frac{13}{15}
\][/tex]
The solutions to the quadratic equation are [tex]\(x = 0\)[/tex] and [tex]\(x = -\frac{13}{15}\)[/tex].
Based on the available options:
- [tex]\( x = -\frac{13}{15}, 0 \)[/tex]
So the correct choice is A.
- [tex]\(a = 15\)[/tex]
- [tex]\(b = 13\)[/tex]
- [tex]\(c = 0\)[/tex]
The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Let's calculate the solutions step-by-step:
1. Calculate the Discriminant:
[tex]\[
b^2 - 4ac = 13^2 - 4 \times 15 \times 0 = 169 - 0 = 169
\][/tex]
2. Solve for [tex]\(x\)[/tex] using the quadratic formula:
[tex]\[
x = \frac{-13 \pm \sqrt{169}}{2 \times 15}
\][/tex]
- Since [tex]\(\sqrt{169} = 13\)[/tex], we compute the two possible values for [tex]\(x\)[/tex]:
- First solution:
[tex]\[
x_1 = \frac{-13 + 13}{30} = \frac{0}{30} = 0
\][/tex]
- Second solution:
[tex]\[
x_2 = \frac{-13 - 13}{30} = \frac{-26}{30} = -\frac{26}{30} = -\frac{13}{15}
\][/tex]
The solutions to the quadratic equation are [tex]\(x = 0\)[/tex] and [tex]\(x = -\frac{13}{15}\)[/tex].
Based on the available options:
- [tex]\( x = -\frac{13}{15}, 0 \)[/tex]
So the correct choice is A.
