Question:
If the quadratic equations
\[
x^2+2x-4k=0
\]
and
\[
x^2+9x+3k=0
\]
have exactly one common root and \(k\neq 0\), then the equation having \(k\) as a root is
If the quadratic equations
\[
x^2+2x-4k=0
\]
and
\[
x^2+9x+3k=0
\]
have exactly one common root and \(k\neq 0\), then the equation having \(k\) as a root is
Show Hint
When two polynomial equations have a common root, subtract the equations first. This usually eliminates the highest-degree terms and makes finding the common root much easier.
Updated On: Jul 29, 2026
- \(x^2+5x-6=0\)
- \(x^2-6x+8=0\)
- \(x^2-5x-6=0\)
- \(x^2-2x-8=0\)
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
If two quadratic equations have a common root \(\alpha\), then \(\alpha\) must satisfy both equations. Subtracting the equations helps determine the common root.
Step 1: Find the common root. Let \(\alpha\) be the common root. Then \[ \alpha^2+2\alpha-4k=0 \] and \[ \alpha^2+9\alpha+3k=0. \] Subtracting, \[ 7\alpha+7k=0. \] \[ \alpha=-k. \]
Step 2: Substitute \(\alpha=-k\) into one equation. Substituting in \[ \alpha^2+2\alpha-4k=0, \] we get \[ k^2-2k-4k=0. \] \[ k^2-6k=0. \] \[ k(k-6)=0. \] Since \(k\neq0\), \[ k=6. \]
Step 3: Find the equation having \(k\) as a root. Substituting \(k=6\) in the options: For option (C), \[ 6^2-5(6)-6 = 36-30-6 = 0. \] Hence \(k=6\) is a root of \[ x^2-5x-6=0. \] Therefore, \[ \boxed{x^2-5x-6=0} \]
Step 1: Find the common root. Let \(\alpha\) be the common root. Then \[ \alpha^2+2\alpha-4k=0 \] and \[ \alpha^2+9\alpha+3k=0. \] Subtracting, \[ 7\alpha+7k=0. \] \[ \alpha=-k. \]
Step 2: Substitute \(\alpha=-k\) into one equation. Substituting in \[ \alpha^2+2\alpha-4k=0, \] we get \[ k^2-2k-4k=0. \] \[ k^2-6k=0. \] \[ k(k-6)=0. \] Since \(k\neq0\), \[ k=6. \]
Step 3: Find the equation having \(k\) as a root. Substituting \(k=6\) in the options: For option (C), \[ 6^2-5(6)-6 = 36-30-6 = 0. \] Hence \(k=6\) is a root of \[ x^2-5x-6=0. \] Therefore, \[ \boxed{x^2-5x-6=0} \]
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