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Answer :
To solve the quadratic equation [tex]\(15x^2 + 13x = 0\)[/tex], we can use the quadratic formula. The general form of a quadratic equation is [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 given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
However, since [tex]\(c = 0\)[/tex], the equation [tex]\(15x^2 + 13x = 0\)[/tex] can be factored easily. We can factor out an [tex]\(x\)[/tex]:
[tex]\[ x(15x + 13) = 0 \][/tex]
This gives us two solutions:
1. [tex]\( x = 0 \)[/tex]
2. [tex]\( 15x + 13 = 0 \)[/tex]
To solve [tex]\(15x + 13 = 0\)[/tex], subtract 13 from both sides:
[tex]\[ 15x = -13 \][/tex]
Divide both sides by 15:
[tex]\[ x = -\frac{13}{15} \][/tex]
So, the solutions are [tex]\(x = 0\)[/tex] and [tex]\(x = -\frac{13}{15}\)[/tex].
Among the options provided:
- a. [tex]\(x = -\frac{13}{15}, 0\)[/tex]
- b. [tex]\(x = 0\)[/tex]
- c. [tex]\(x = \frac{13}{15}, 0\)[/tex]
- d. [tex]\(x = \pm \frac{13}{15}\)[/tex]
The correct choice for the solutions is option A: [tex]\(x = -\frac{13}{15}, 0\)[/tex].
- [tex]\(a = 15\)[/tex]
- [tex]\(b = 13\)[/tex]
- [tex]\(c = 0\)[/tex]
The quadratic formula is given by:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
However, since [tex]\(c = 0\)[/tex], the equation [tex]\(15x^2 + 13x = 0\)[/tex] can be factored easily. We can factor out an [tex]\(x\)[/tex]:
[tex]\[ x(15x + 13) = 0 \][/tex]
This gives us two solutions:
1. [tex]\( x = 0 \)[/tex]
2. [tex]\( 15x + 13 = 0 \)[/tex]
To solve [tex]\(15x + 13 = 0\)[/tex], subtract 13 from both sides:
[tex]\[ 15x = -13 \][/tex]
Divide both sides by 15:
[tex]\[ x = -\frac{13}{15} \][/tex]
So, the solutions are [tex]\(x = 0\)[/tex] and [tex]\(x = -\frac{13}{15}\)[/tex].
Among the options provided:
- a. [tex]\(x = -\frac{13}{15}, 0\)[/tex]
- b. [tex]\(x = 0\)[/tex]
- c. [tex]\(x = \frac{13}{15}, 0\)[/tex]
- d. [tex]\(x = \pm \frac{13}{15}\)[/tex]
The correct choice for the solutions is option A: [tex]\(x = -\frac{13}{15}, 0\)[/tex].
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