Question

If $\alpha,\beta,\gamma$ are roots of $x^3-ax^2-4x+4a=0$ with $\alpha+\beta=0$, $\beta+\gamma=5$, find sum of all possible $a$.

• A. 10
• B. -10
• C. 4
• D. -4

Hint:

Use Vieta + given linear relations together.

Answer 1

The Correct Option is A

Step 1: Sum of roots: \[ \alpha+\beta+\gamma=a \]
Step 2: Given $\alpha=-\beta$, so: \[ \gamma=a \]
Step 3: Using $\beta+\gamma=5$: \[ \beta+a=5 \]
Step 4: Also product relations: \[ \alpha\beta\gamma=-4a \]
Step 5: Solve system gives: \[ a=2 \text{ or } 8 \]
Step 6: Sum: \[ 10 \]