If ` alpha , beta , gamma ` are the roots of ` x^3 +px^2 +qx +r=0` then find
`sum alpha^2`

Answer

To find the sum of the squares of the roots \( \alpha^2 + \beta^2 + \gamma^2 \) for the polynomial equation \( x^3 + px^2 + qx + r = 0 \), we can follow these steps:

### Step 1: Identify the coefficients
The given polynomial is:
\[ x^3 + px^2 + qx + r = 0 \]
Here, we can identify the coefficients:
- \( a = 1 \)
- \( b = p \)
- \( c = q \)
- \( d = r \)

### Step 2: Use Vieta's formulas
According to Vieta's formulas for a cubic equation \( ax^3 + bx^2 + cx + d = 0 \):
- The sum of the roots \( \alpha + \beta + \gamma = -\frac{b}{a} = -p \)
- The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = q \)
- The product of the roots \( \alpha\beta\gamma = -\frac{d}{a} = -r \)

### Step 3: Find the sum of the squares of the roots
We need to find \( \alpha^2 + \beta^2 + \gamma^2 \). We can use the identity:
\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \]

### Step 4: Substitute the values from Vieta's formulas
Substituting the values we found using Vieta's formulas:
\[ \alpha^2 + \beta^2 + \gamma^2 = (-p)^2 - 2q \]

### Step 5: Simplify the expression
Now, simplifying the expression gives us:
\[ \alpha^2 + \beta^2 + \gamma^2 = p^2 - 2q \]

### Final Answer
Thus, the sum of the squares of the roots is:
\[ \alpha^2 + \beta^2 + \gamma^2 = p^2 - 2q \]

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