If \(x=-1\) is a root of \(x^2+(6k-3)x+5k=0\), what is the value of \(k\)?
Answer and explanation
Correct answer: 4
Substitute \(x=-1\): \(1+(6k-3)(-1)+5k=0\) which simplifies to \(1-6k+3+5k=0\), i.e. \(4-k=0\). Therefore \(k=4\). Option B (\(-4\)) often results from a sign error during substitution; the correct simplification yields a positive 4. Exam tip: Always substitute the root directly and simplify terms carefully, watching for sign changes when \(x\) is negative.
Frequently asked questions
What is the correct answer to this question?
4
Why is this the correct answer?
Substitute \(x=-1\): \(1+(6k-3)(-1)+5k=0\) which simplifies to \(1-6k+3+5k=0\), i.e. \(4-k=0\). Therefore \(k=4\). Option B (\(-4\)) often results from a sign error during substitution; the correct simplification yields a positive 4. Exam tip: Always substitute the root directly and simplify terms carefully, watching for sign changes when \(x\) is negative.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Introduction to Quadratic Equations.
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