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Answer :
To solve the equation [tex]\(15x^2 + 13x = 0\)[/tex] using the quadratic formula, let's go through the steps systematically:
1. Identify the coefficients:
The given equation is already in the form [tex]\(ax^2 + bx + c = 0\)[/tex]. Here:
- [tex]\(a = 15\)[/tex]
- [tex]\(b = 13\)[/tex]
- [tex]\(c = 0\)[/tex]
2. Quadratic Formula:
The quadratic formula is used to find the solutions of a quadratic equation and is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
3. Calculate the Discriminant:
The discriminant [tex]\(\Delta\)[/tex] is calculated as:
[tex]\[
\Delta = b^2 - 4ac = 13^2 - 4 \times 15 \times 0 = 169 - 0 = 169
\][/tex]
4. Find the Solutions:
Using the quadratic formula, we substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(\Delta\)[/tex]:
[tex]\[
x = \frac{-13 \pm \sqrt{169}}{2 \times 15}
\][/tex]
[tex]\[
x = \frac{-13 \pm 13}{30}
\][/tex]
- First solution:
[tex]\[
x_1 = \frac{-13 + 13}{30} = \frac{0}{30} = 0
\][/tex]
- Second solution:
[tex]\[
x_2 = \frac{-13 - 13}{30} = \frac{-26}{30} = -\frac{13}{15}
\][/tex]
5. Write the Solutions:
The solutions to the equation [tex]\(15x^2 + 13x = 0\)[/tex] are [tex]\(x = 0\)[/tex] and [tex]\(x = -\frac{13}{15}\)[/tex].
Given the options, the correct answer is:
- a. [tex]\(x = -\frac{13}{15}, 0\)[/tex]
So, option A is the correct choice.
1. Identify the coefficients:
The given equation is already in the form [tex]\(ax^2 + bx + c = 0\)[/tex]. Here:
- [tex]\(a = 15\)[/tex]
- [tex]\(b = 13\)[/tex]
- [tex]\(c = 0\)[/tex]
2. Quadratic Formula:
The quadratic formula is used to find the solutions of a quadratic equation and is given by:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
3. Calculate the Discriminant:
The discriminant [tex]\(\Delta\)[/tex] is calculated as:
[tex]\[
\Delta = b^2 - 4ac = 13^2 - 4 \times 15 \times 0 = 169 - 0 = 169
\][/tex]
4. Find the Solutions:
Using the quadratic formula, we substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(\Delta\)[/tex]:
[tex]\[
x = \frac{-13 \pm \sqrt{169}}{2 \times 15}
\][/tex]
[tex]\[
x = \frac{-13 \pm 13}{30}
\][/tex]
- First solution:
[tex]\[
x_1 = \frac{-13 + 13}{30} = \frac{0}{30} = 0
\][/tex]
- Second solution:
[tex]\[
x_2 = \frac{-13 - 13}{30} = \frac{-26}{30} = -\frac{13}{15}
\][/tex]
5. Write the Solutions:
The solutions to the equation [tex]\(15x^2 + 13x = 0\)[/tex] are [tex]\(x = 0\)[/tex] and [tex]\(x = -\frac{13}{15}\)[/tex].
Given the options, the correct answer is:
- a. [tex]\(x = -\frac{13}{15}, 0\)[/tex]
So, option A is the correct choice.
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