Question

If \(-5\) and \(-1\) are the roots of a quadratic equation, then the equation will be:

• A. \(x^2 + 6x + 5 = 0\)
• B. \(x^2 - 6x + 5 = 0\)
• C. \(x^2 - 6x - 5 = 0\)
• D. \(x^2 + 6x - 5 = 0\)

Hint:

Given roots \(\alpha,\beta\), the monic quadratic is \(x^2-(\alpha+\beta)x+\alpha\beta=0\). This avoids expanding factors each time.

Answer 1

The Correct Option is A

Step 1: Use sum and product of roots for a monic quadratic.
If roots are \( \alpha = -5 \) and \( \beta = -1 \), then
\[ \alpha+\beta = -6, \quad \alpha\beta = 5. \]
Step 2: Form the quadratic using \(x^2 - (\alpha+\beta)x + \alpha\beta = 0\).
\[ x^2 - (-6)x + 5 = 0 \;\Rightarrow\; x^2 + 6x + 5 = 0. \]
Step 3: Conclude.
Hence, the required equation is \( x^2 + 6x + 5 = 0 \).