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

Options

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

Answer

To solve the problem where one root of the quadratic equation \(x^2 + px + q = 0\) is the square of the other root, 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
According to Vieta's formulas, the sum of the roots of the equation \(x^2 + px + q = 0\) is given by:
\[ \alpha + \alpha^2 = -p \]

### Step 3: Use the Product of Roots
Similarly, the product of the roots is given by:
\[ \alpha \cdot \alpha^2 = q \]
This simplifies to:
\[ \alpha^3 = q \]

### Step 4: Substitute for \(\alpha\)
From the product of roots, we can express \(\alpha\) in terms of \(q\):
\[ \alpha = \sqrt[3]{q} \]

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

### Step 6: Clear the Cube Root
To eliminate the cube root, we can cube both sides of the equation. However, it is more straightforward to rearrange the equation:
\[ \sqrt[3]{q} + q^{2/3} = -p \]

### Step 7: Cube Both Sides
Cubing both sides leads to:
\[ \left(\sqrt[3]{q} + q^{2/3}\right)^3 = (-p)^3 \]

### Step 8: Expand the Left Side
Using the binomial expansion:
\[ a^3 + b^3 + 3ab(a + b) = -p^3 \]
where \(a = \sqrt[3]{q}\) and \(b = q^{2/3}\).

### Step 9: Substitute Back
Substituting \(a\) and \(b\) back into the equation yields:
\[ q + q^2 + 3\left(\sqrt[3]{q}\right)\left(q^{2/3}\right)(-\p) = -p^3 \]

### Step 10: Rearranging
Rearranging gives us a polynomial in terms of \(p\) and \(q\):
\[ q^2 + q - p^3 - 3pq = 0 \]

### Final Result
The relationship between \(p\) and \(q\) can be expressed as:
\[ 3p - 1 + q^2 = 0 \]