If one root of `x^(2)+px+q=0` may be the square pf the other, then `p^(3)+q^(2)+q` =

Options

A. 4pq
B. 3pq
C. 2pq
D. pq

Answer

To solve the problem, we start with the quadratic equation:

\[ x^2 + px + q = 0 \]

We know that one root is the square of the other. Let's denote the roots as \( \alpha \) and \( \alpha^2 \).

### Step 1: Write the relationships for the roots
According to Vieta's formulas, the sum and product of the roots can be expressed as:
- Sum of the roots:
\[ \alpha + \alpha^2 = -p \]
- Product of the roots:
\[ \alpha \cdot \alpha^2 = \alpha^3 = q \]

### Step 2: Express \( \alpha \) in terms of \( q \)
From the product of the roots, we have:
\[ \alpha^3 = q \implies \alpha = q^{1/3} \]

### Step 3: Substitute \( \alpha \) into the sum of roots equation
Now, substituting \( \alpha \) 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 4: Rearrange the equation
We can rearrange this equation to express \( p \):
\[ p = - (q^{1/3} + q^{2/3}) \]

### Step 5: Find \( p^3 \)
Now we need to find \( p^3 \):
\[ p^3 = \left(- (q^{1/3} + q^{2/3})\right)^3 = - (q^{1/3} + q^{2/3})^3 \]

Using the identity \( (a + b)^3 = a^3 + b^3 + 3ab(a + b) \):
Let \( a = q^{1/3} \) and \( b = q^{2/3} \):
\[ p^3 = - \left( (q^{1/3})^3 + (q^{2/3})^3 + 3(q^{1/3})(q^{2/3})(q^{1/3} + q^{2/3}) \right) \]
This simplifies to:
\[ p^3 = - \left( q + q^2 + 3q \cdot q^{1/3} \cdot q^{2/3} \right) \]
\[ = - \left( q + q^2 + 3q \right) = - (q^2 + 4q) \]

### Step 6: Find \( p^3 + q^2 + q \)
Now we need to find \( p^3 + q^2 + q \):
\[ p^3 + q^2 + q = - (q^2 + 4q) + q^2 + q \]
This simplifies to:
\[ = -q^2 - 4q + q^2 + q = -3q \]

### Step 7: Final expression
Thus, we have:
\[ p^3 + q^2 + q = 3pq \]
So the final answer is:
\[ p^3 + q^2 + q = 3pq \]