High School

Solve the equation using the quadratic formula:

[tex]15x^2 + 13x = 0[/tex]

A. [tex]x = -\frac{13}{15}, 0[/tex]
B. [tex]x = \frac{13}{15}, 0[/tex]
C. [tex]x = 0[/tex]
D. [tex]x = \pm \frac{13}{15}[/tex]

Answer :

We start with the quadratic equation

[tex]$$15x^2 + 13x = 0.$$[/tex]

### Step 1. Identify the coefficients
In the standard form of a quadratic equation, [tex]$ax^2 + bx + c = 0$[/tex], we have:

[tex]$$a = 15,\quad b = 13,\quad c = 0.$$[/tex]

### Step 2. Compute the discriminant
The discriminant is given by

[tex]$$D = b^2 - 4ac.$$[/tex]

Substitute the values:

[tex]$$D = 13^2 - 4(15)(0) = 169 - 0 = 169.$$[/tex]

### Step 3. Compute the square root of the discriminant
Since

[tex]$$\sqrt{D} = \sqrt{169} = 13,$$[/tex]

we now have the necessary value to apply in the quadratic formula.

### Step 4. Apply the quadratic formula
The quadratic formula states

[tex]$$x = \frac{-b \pm \sqrt{D}}{2a}.$$[/tex]

Substitute the known values:

[tex]$$x = \frac{-13 \pm 13}{2 \times 15} = \frac{-13 \pm 13}{30}.$$[/tex]

### Step 5. Solve for the two values of [tex]$x$[/tex]
1. For the positive square root:

[tex]$$x = \frac{-13 + 13}{30} = \frac{0}{30} = 0.$$[/tex]

2. For the negative square root:

[tex]$$x = \frac{-13 - 13}{30} = \frac{-26}{30} = -\frac{13}{15}.$$[/tex]

### Final Answer
The solutions to the equation are

[tex]$$x = 0 \quad \text{and} \quad x = -\frac{13}{15}.$$[/tex]

Thus, the correct answer is:

[tex]$$x = -\frac{13}{15},\ 0.$$[/tex]