Find the condition that one root of the quadratic equation `ax^(2)+bx+c=0` shall be n times the other, where n is positive integer.

Options

A. 1) `nb^2`=`ac(1+n)^2`
B. 2)`nb`=`ac(1+n)^2`
C. 3)`nb^2`=`ac(1+n)`
D. 4)none of these

Answer

To find the condition that one root of the quadratic equation \( ax^2 + bx + c = 0 \) shall be \( n \) times the other, where \( n \) is a positive integer, we can follow these steps:

### Step-by-Step Solution:

1. **Assume the Roots**:
Let the first root be \( p \) and the second root be \( np \) (since one root is \( n \) times the other).

2. **Sum of the Roots**:
According to Vieta's formulas, the sum of the roots of the quadratic equation is given by:
\[ p + np = -\frac{b}{a} \]
This can be factored as:
\[ p(1 + n) = -\frac{b}{a} \]
From this, we can express \( p \) as:
\[ p = -\frac{b}{a(1+n)} \]

3. **Product of the Roots**:
The product of the roots is also given by Vieta's formulas:
\[ p \cdot np = \frac{c}{a} \]
This simplifies to:
\[ np^2 = \frac{c}{a} \]

4. **Substituting \( p \)**:
Now, substitute the value of \( p \) from step 2 into the product equation:
\[ n \left(-\frac{b}{a(1+n)}\right)^2 = \frac{c}{a} \]
Simplifying the left-hand side:
\[ n \cdot \frac{b^2}{a^2(1+n)^2} = \frac{c}{a} \]

5. **Cross-Multiplying**:
Cross-multiply to eliminate the fractions:
\[ n b^2 = c \cdot a(1+n)^2 \]

6. **Final Condition**:
Thus, the condition that one root is \( n \) times the other is:
\[ n b^2 = ac(1+n)^2 \]

### Final Answer:
The condition is:
\[ n b^2 = ac(1+n)^2 \]