Answer :
To solve the equation [tex]\(15x^2 + 13x = 0\)[/tex] using the quadratic formula, we first identify the coefficients from the standard form of a quadratic equation, which is [tex]\(ax^2 + bx + c = 0\)[/tex]. Here, [tex]\(a = 15\)[/tex], [tex]\(b = 13\)[/tex], and [tex]\(c = 0\)[/tex].
The quadratic formula is given by:
[tex]\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
\][/tex]
Let's apply these values to the formula:
1. Calculate the Discriminant:
[tex]\[
b^2 - 4ac = 13^2 - 4 \times 15 \times 0 = 169 - 0 = 169
\][/tex]
2. Find the Roots:
Since the discriminant is [tex]\(169\)[/tex], which is a perfect square, we will find two real roots:
- The first root ([tex]\(x_1\)[/tex]) is:
[tex]\[
x_1 = \frac{{-b + \sqrt{{b^2 - 4ac}}}}{2a} = \frac{{-13 + \sqrt{169}}}{2 \times 15}
\][/tex]
[tex]\[
x_1 = \frac{{-13 + 13}}{30} = \frac{0}{30} = 0
\][/tex]
- The second root ([tex]\(x_2\)[/tex]) is:
[tex]\[
x_2 = \frac{{-b - \sqrt{{b^2 - 4ac}}}}{2a} = \frac{{-13 - \sqrt{169}}}{2 \times 15}
\][/tex]
[tex]\[
x_2 = \frac{{-13 - 13}}{30} = \frac{-26}{30} = -\frac{13}{15}
\][/tex]
Therefore, the solutions are [tex]\(x = 0\)[/tex] and [tex]\(x = -\frac{13}{15}\)[/tex].
From the provided answer choices:
- a. [tex]\(x = -\frac{13}{15}, 0\)[/tex]
- b. [tex]\(x = 0\)[/tex]
- c. [tex]\(x = \frac{13}{15}, 0\)[/tex]
The correct choice is A: [tex]\(x = -\frac{13}{15}, 0\)[/tex].
The quadratic formula is given by:
[tex]\[
x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}
\][/tex]
Let's apply these values to the formula:
1. Calculate the Discriminant:
[tex]\[
b^2 - 4ac = 13^2 - 4 \times 15 \times 0 = 169 - 0 = 169
\][/tex]
2. Find the Roots:
Since the discriminant is [tex]\(169\)[/tex], which is a perfect square, we will find two real roots:
- The first root ([tex]\(x_1\)[/tex]) is:
[tex]\[
x_1 = \frac{{-b + \sqrt{{b^2 - 4ac}}}}{2a} = \frac{{-13 + \sqrt{169}}}{2 \times 15}
\][/tex]
[tex]\[
x_1 = \frac{{-13 + 13}}{30} = \frac{0}{30} = 0
\][/tex]
- The second root ([tex]\(x_2\)[/tex]) is:
[tex]\[
x_2 = \frac{{-b - \sqrt{{b^2 - 4ac}}}}{2a} = \frac{{-13 - \sqrt{169}}}{2 \times 15}
\][/tex]
[tex]\[
x_2 = \frac{{-13 - 13}}{30} = \frac{-26}{30} = -\frac{13}{15}
\][/tex]
Therefore, the solutions are [tex]\(x = 0\)[/tex] and [tex]\(x = -\frac{13}{15}\)[/tex].
From the provided answer choices:
- a. [tex]\(x = -\frac{13}{15}, 0\)[/tex]
- b. [tex]\(x = 0\)[/tex]
- c. [tex]\(x = \frac{13}{15}, 0\)[/tex]
The correct choice is A: [tex]\(x = -\frac{13}{15}, 0\)[/tex].
