Answer :
To solve the equation [tex]\(15x^2 + 13x = 0\)[/tex] using the quadratic formula, we need to find the values of [tex]\(x\)[/tex] that satisfy this equation. The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are the coefficients from the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex].
For the given equation [tex]\(15x^2 + 13x = 0\)[/tex], we have:
- [tex]\(a = 15\)[/tex]
- [tex]\(b = 13\)[/tex]
- [tex]\(c = 0\)[/tex]
Now, substitute these values into the formula:
1. Find the discriminant:
[tex]\[ b^2 - 4ac = 13^2 - 4 \times 15 \times 0 = 169 \][/tex]
2. Calculate the roots using the quadratic formula:
[tex]\[ x_1 = \frac{-13 - \sqrt{169}}{30} \][/tex]
[tex]\[ x_1 = \frac{-13 - 13}{30} \][/tex]
[tex]\[ x_1 = \frac{-26}{30} \][/tex]
[tex]\[ x_1 = -\frac{13}{15} \][/tex]
[tex]\[ x_2 = \frac{-13 + \sqrt{169}}{30} \][/tex]
[tex]\[ x_2 = \frac{-13 + 13}{30} \][/tex]
[tex]\[ x_2 = \frac{0}{30} \][/tex]
[tex]\[ x_2 = 0 \][/tex]
Thus, the solutions are [tex]\(x = -\frac{13}{15}\)[/tex] and [tex]\(x = 0\)[/tex].
The best answer choice that matches these solutions is:
A. [tex]\(x = -\frac{13}{15}, 0\)[/tex]
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] are the coefficients from the quadratic equation [tex]\(ax^2 + bx + c = 0\)[/tex].
For the given equation [tex]\(15x^2 + 13x = 0\)[/tex], we have:
- [tex]\(a = 15\)[/tex]
- [tex]\(b = 13\)[/tex]
- [tex]\(c = 0\)[/tex]
Now, substitute these values into the formula:
1. Find the discriminant:
[tex]\[ b^2 - 4ac = 13^2 - 4 \times 15 \times 0 = 169 \][/tex]
2. Calculate the roots using the quadratic formula:
[tex]\[ x_1 = \frac{-13 - \sqrt{169}}{30} \][/tex]
[tex]\[ x_1 = \frac{-13 - 13}{30} \][/tex]
[tex]\[ x_1 = \frac{-26}{30} \][/tex]
[tex]\[ x_1 = -\frac{13}{15} \][/tex]
[tex]\[ x_2 = \frac{-13 + \sqrt{169}}{30} \][/tex]
[tex]\[ x_2 = \frac{-13 + 13}{30} \][/tex]
[tex]\[ x_2 = \frac{0}{30} \][/tex]
[tex]\[ x_2 = 0 \][/tex]
Thus, the solutions are [tex]\(x = -\frac{13}{15}\)[/tex] and [tex]\(x = 0\)[/tex].
The best answer choice that matches these solutions is:
A. [tex]\(x = -\frac{13}{15}, 0\)[/tex]
