Question:
If the equation \(x^2 + px + q = 0\) and \(x^2 + qx + p = 0\) have a common root then \(p + q + 1\) is equal to
If the equation \(x^2 + px + q = 0\) and \(x^2 + qx + p = 0\) have a common root then \(p + q + 1\) is equal to
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If two quadratics have a common root, subtract to find the root.
Updated On: Apr 7, 2026
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept:
Let \(\alpha\) be common root, then subtract equations.
Step 2: Detailed Explanation:
\(\alpha^2 + p\alpha + q = 0\) ... (1)
\(\alpha^2 + q\alpha + p = 0\) ... (2)
Subtract: \((p - q)\alpha + (q - p) = 0 \rightarrow (p - q)(\alpha - 1) = 0\)
If \(p \neq q\), then \(\alpha = 1\)
Substitute in (1): \(1 + p + q = 0 \rightarrow p + q + 1 = 0\)
If \(p = q\), then equations identical, any common root. Then \(p = q\).
Step 3: Final Answer:
\(p + q + 1 = 0\).
Let \(\alpha\) be common root, then subtract equations.
Step 2: Detailed Explanation:
\(\alpha^2 + p\alpha + q = 0\) ... (1)
\(\alpha^2 + q\alpha + p = 0\) ... (2)
Subtract: \((p - q)\alpha + (q - p) = 0 \rightarrow (p - q)(\alpha - 1) = 0\)
If \(p \neq q\), then \(\alpha = 1\)
Substitute in (1): \(1 + p + q = 0 \rightarrow p + q + 1 = 0\)
If \(p = q\), then equations identical, any common root. Then \(p = q\).
Step 3: Final Answer:
\(p + q + 1 = 0\).
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