If one root of the equation `x^2+px+q=0` is the square of the other then

Options

A. `p^(3)+q^(2)-q(3p+1)=0`
B. `p^(3)+q^(2)+q(1+3p)=0`
C. `p^(3)+q^(2)+q(3p-1)=0`
D. `p^(3)+q^(2)+q(1-3p)=0`

Answer

To solve the problem where one root of the equation \(x^2 + px + q = 0\) is the square of the other, we can follow these steps:

### Step 1: Define the Roots
Let the roots of the equation be \(\alpha\) and \(\alpha^2\), where \(\alpha\) is one root and \(\alpha^2\) is the square of the other root.

### Step 2: Use the Sum of Roots Formula
According to Vieta's formulas, the sum of the roots of the quadratic equation \(ax^2 + bx + c = 0\) is given by:
\[ \text{Sum of roots} = -\frac{b}{a} \]
For our equation \(x^2 + px + q = 0\), we have:
\[ \alpha + \alpha^2 = -p \]

### Step 3: Use the Product of Roots Formula
The product of the roots is given by:
\[ \text{Product of roots} = \frac{c}{a} \]
For our equation, this gives:
\[ \alpha \cdot \alpha^2 = q \]
This simplifies to:
\[ \alpha^3 = q \]

### Step 4: Express \(\alpha\) in terms of \(q\)
From the equation \(\alpha^3 = q\), we can express \(\alpha\) as:
\[ \alpha = q^{1/3} \]

### Step 5: Substitute \(\alpha\) into the Sum of Roots Equation
Now substitute \(\alpha = q^{1/3}\) into the sum of roots equation:
\[ q^{1/3} + (q^{1/3})^2 = -p \]
This simplifies to:
\[ q^{1/3} + q^{2/3} = -p \]

### Step 6: Rearranging the Equation
We can rearrange the equation:
\[ q^{1/3} + q^{2/3} + p = 0 \]
This can be rewritten as:
\[ q^{1/3} + q^{2/3} = -p \]

### Step 7: Cube Both Sides
To eliminate the roots, we can cube both sides:
\[ (q^{1/3} + q^{2/3})^3 = (-p)^3 \]
Using the identity \((a + b)^3 = a^3 + b^3 + 3ab(a + b)\), we have:
\[ q + q^2 + 3(q^{1/3})(q^{2/3})(q^{1/3} + q^{2/3}) = -p^3 \]

### Step 8: Substitute Back
Substituting back \(q^{1/3} + q^{2/3} = -p\):
\[ q + q^2 - 3p(-p) = -p^3 \]
This simplifies to:
\[ q + q^2 + 3p^2 = -p^3 \]

### Step 9: Final Rearrangement
Rearranging gives:
\[ q + q^2 = 2p \]

### Conclusion
Thus, we conclude that if one root of the equation \(x^2 + px + q = 0\) is the square of the other, we have:
\[ q + q^2 = 2p \]