The roots of the equation `x^(2)+px+q=0` are 1 and 2 .The roots of the equation `qx^(2)-px+1=0` will be

Options

A. `-(1)/(2),-1`
B. `(1)/(2),1`
C. `-(1)/(2),1`
D. None of these

Answer

To solve the problem step by step, we need to find the roots of the equation \( qx^2 - px + 1 = 0 \) given that the roots of the equation \( x^2 + px + q = 0 \) are 1 and 2.

### Step 1: Find the values of \( p \) and \( q \)

Given that the roots of the equation \( x^2 + px + q = 0 \) are 1 and 2, we can use the relationships for the sum and product of the roots.

- **Sum of the roots**: \( r_1 + r_2 = 1 + 2 = 3 \)
- **Product of the roots**: \( r_1 \cdot r_2 = 1 \cdot 2 = 2 \)

From Vieta's formulas, we know:

1. The sum of the roots is given by \( -p \) (since the equation is in the form \( ax^2 + bx + c = 0 \)).
\[ -p = 3 \implies p = -3 \]

2. The product of the roots is given by \( q \).
\[ q = 2 \]

### Step 2: Substitute \( p \) and \( q \) into the new equation

Now that we have \( p = -3 \) and \( q = 2 \), we can substitute these values into the new equation \( qx^2 - px + 1 = 0 \):

\[ 2x^2 - (-3)x + 1 = 0 \implies 2x^2 + 3x + 1 = 0 \]

### Step 3: Factor or use the quadratic formula to find the roots

We can factor the quadratic equation \( 2x^2 + 3x + 1 = 0 \). We look for two numbers that multiply to \( 2 \cdot 1 = 2 \) and add to \( 3 \).

The factors of \( 2 \) that add up to \( 3 \) are \( 2 \) and \( 1 \). Therefore, we can factor the equation as follows:

\[ 2x^2 + 2x + x + 1 = 0 \]
\[ 2x(x + 1) + 1(x + 1) = 0 \]
\[ (2x + 1)(x + 1) = 0 \]

### Step 4: Set each factor to zero to find the roots

Setting each factor to zero gives us:

1. \( 2x + 1 = 0 \)
\[ 2x = -1 \implies x = -\frac{1}{2} \]

2. \( x + 1 = 0 \)
\[ x = -1 \]

### Conclusion

The roots of the equation \( qx^2 - px + 1 = 0 \) are \( x = -\frac{1}{2} \) and \( x = -1 \).