If the two roots of a quadratic equation are `(-3)/2` and `(-5)/3` which one of the following is the concerned quadratic equation?

Options

A. `6x^2+19x+15`
B. `199x+6x^2+15`
C. `30x^2+95x+75`
D. both (a) and (c)

Answer

To find the quadratic equation given its roots, we can use the relationship between the roots and the coefficients of the quadratic equation. The general form of a quadratic equation is:

\[ ax^2 + bx + c = 0 \]

where \( a \), \( b \), and \( c \) are constants. If the roots of the equation are \( r_1 \) and \( r_2 \), then we can express the equation as:

\[ x^2 - (r_1 + r_2)x + (r_1 \cdot r_2) = 0 \]

### Step-by-step Solution:

1. **Identify the roots**: The roots given are \( r_1 = -\frac{3}{2} \) and \( r_2 = -\frac{5}{3} \).

2. **Calculate the sum of the roots**:
\[ r_1 + r_2 = -\frac{3}{2} + -\frac{5}{3} \]
To add these fractions, we need a common denominator. The least common multiple of 2 and 3 is 6.
\[ r_1 + r_2 = -\frac{3 \cdot 3}{2 \cdot 3} - \frac{5 \cdot 2}{3 \cdot 2} = -\frac{9}{6} - \frac{10}{6} = -\frac{19}{6} \]

3. **Calculate the product of the roots**:
\[ r_1 \cdot r_2 = -\frac{3}{2} \cdot -\frac{5}{3} = \frac{15}{6} = \frac{5}{2} \]

4. **Form the quadratic equation**:
Using the sum and product of the roots, we can write the quadratic equation:
\[ x^2 - (r_1 + r_2)x + (r_1 \cdot r_2) = 0 \]
Substituting the values we found:
\[ x^2 - \left(-\frac{19}{6}\right)x + \frac{5}{2} = 0 \]
This simplifies to:
\[ x^2 + \frac{19}{6}x + \frac{5}{2} = 0 \]

5. **Clear the fractions**:
To eliminate the fractions, multiply the entire equation by 6 (the least common multiple of the denominators):
\[ 6x^2 + 19x + 15 = 0 \]

### Final Quadratic Equation:
The concerned quadratic equation is:
\[ 6x^2 + 19x + 15 = 0 \]