Question

If \( \alpha, \beta \) are the roots of the equation \( x^2 + 3px + 2p^2 = 0 \) and \( \alpha^2 + \beta^2 = 5 \), then the value of \(p\) is:

• A. \( \pm 3 \)
• B. \( \pm 2 \)
• C. \( \pm 1 \)
• D. \( \pm 5 \)

Hint:

For a monic quadratic \(x^2+bx+c=0\), use Vieta: \(\alpha+\beta=-b\), \(\alpha\beta=c\). Also, \(\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta\).

Answer 1

The Correct Option is C

Step 1: Use Vieta's formulas for the quadratic \(x^2 + 3px + 2p^2 = 0\).
Sum of roots: \( \alpha + \beta = -3p \).
Product of roots: \( \alpha\beta = 2p^2 \).
Step 2: Express \( \alpha^2 + \beta^2 \) in terms of \(p\).
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (-3p)^2 - 2(2p^2) = 9p^2 - 4p^2 = 5p^2. \]
Step 3: Use the given condition \( \alpha^2 + \beta^2 = 5 \).
\[ 5p^2 = 5 \;\Rightarrow\; p^2 = 1 \;\Rightarrow\; p = \pm 1. \]
Step 4: Conclude.
Hence, the required values of \(p\) are \( \boxed{\pm 1} \).