The roots of `Ax^(2)+ Bx + C=0` are r and s. For the roots of `x^(2) + px+ q=0 " to be " r^(2) and s^(2),` what must be the value of p?

Options

A. ` (B^(2) - 4AC)//A^(2)`
B. ` (B^(2) - 2AC)//A^(2)`
C. ` (2AC- B^(2))//A^(2)`
D. ` B^(2) - 2C`

Explanation

To find the value of \( p \) such that the roots of the quadratic equation \( x^2 + px + q = 0 \) are \( r^2 \) and \( s^2 \), where \( r \) and \( s \) are the roots of the equation \( Ax^2 + Bx + C = 0 \), we can follow these steps:

### Step 1: Understand the relationships between the roots and coefficients
From Vieta's formulas, we know that for the quadratic equation \( Ax^2 + Bx + C = 0 \):
- The sum of the roots \( r + s = -\frac{B}{A} \)
- The product of the roots \( rs = \frac{C}{A} \)

### Step 2: Express the sum of the new roots
For the new quadratic equation \( x^2 + px + q = 0 \) with roots \( r^2 \) and \( s^2 \):
- The sum of the roots \( r^2 + s^2 \) can be expressed using the identity:
\[ r^2 + s^2 = (r + s)^2 - 2rs \]

### Step 3: Substitute the values from Step 1
Substituting the values of \( r + s \) and \( rs \) from Step 1 into the equation from Step 2:
\[ r^2 + s^2 = \left(-\frac{B}{A}\right)^2 - 2\left(\frac{C}{A}\right) \]
This simplifies to:
\[ r^2 + s^2 = \frac{B^2}{A^2} - \frac{2C}{A} \]

### Step 4: Relate the sum of the new roots to \( p \)
According to Vieta's formulas for the new quadratic equation, the sum of the roots \( r^2 + s^2 \) is equal to \( -p \):
\[ -p = \frac{B^2}{A^2} - \frac{2C}{A} \]

### Step 5: Solve for \( p \)
Rearranging the equation gives:
\[ p = -\left(\frac{B^2}{A^2} - \frac{2C}{A}\right) \]
This can be rewritten as:
\[ p = \frac{2C}{A} - \frac{B^2}{A^2} \]
Multiplying through by \( A^2 \) to eliminate the denominators:
\[ p = \frac{2AC - B^2}{A^2} \]

### Final Result
Thus, the value of \( p \) is:
\[ p = \frac{2AC - B^2}{A^2} \]