If the two roots of a quadratic equation are 7/2 and 3/5 which one of the following is the concerned quadratic equation?

Options

A. `10x^2-41x+21=0`
B. `10x^2+41x-21=0`
C. `5x^2-40x+24=0`
D. `10x^2-24x+45=0`

Answer

To find the quadratic equation given the roots \( \frac{7}{2} \) and \( \frac{3}{5} \), we will follow these steps:

### Step 1: Identify the roots
Let the roots be:
- \( A = \frac{7}{2} \)
- \( B = \frac{3}{5} \)

### Step 2: Calculate the sum of the roots
The sum of the roots \( S \) is given by:
\[ S = A + B = \frac{7}{2} + \frac{3}{5} \]
To add these fractions, we need a common denominator. The least common multiple of 2 and 5 is 10.

Convert each fraction:
\[ \frac{7}{2} = \frac{7 \times 5}{2 \times 5} = \frac{35}{10} \]
\[ \frac{3}{5} = \frac{3 \times 2}{5 \times 2} = \frac{6}{10} \]

Now, add them:
\[ S = \frac{35}{10} + \frac{6}{10} = \frac{41}{10} \]

### Step 3: Calculate the product of the roots
The product of the roots \( P \) is given by:
\[ P = A \times B = \frac{7}{2} \times \frac{3}{5} = \frac{21}{10} \]

### Step 4: Write the quadratic equation
The general form of a quadratic equation with roots \( A \) and \( B \) is:
\[ x^2 - (A + B)x + (A \cdot B) = 0 \]
Substituting the values we calculated:
\[ x^2 - \frac{41}{10}x + \frac{21}{10} = 0 \]

### Step 5: Eliminate the fraction by multiplying through by 10
To eliminate the fractions, multiply the entire equation by 10:
\[ 10x^2 - 41x + 21 = 0 \]

### Conclusion
The required quadratic equation is:
\[ 10x^2 - 41x + 21 = 0 \]