If the two roots of the quadratic equation \(x^2-6x+k=0\) are real and equal, what is the value of \(k\)?
Answer and explanation
Correct answer: 9
For real and equal roots, the discriminant must be zero. Here, \(a=1, b=-6, c=k\), so \(D=b^2-4ac=(-6)^2-4(1)(k)=36-4k\). Setting \(D=0\) gives \(36-4k=0\), hence \(k=9\). The value 6 is only the magnitude of the coefficient of \(x\), so it is not correct. Exam tip: For equal roots of a quadratic equation, always use \(b^2-4ac=0\).
Frequently asked questions
What is the correct answer to this question?
9
Why is this the correct answer?
For real and equal roots, the discriminant must be zero. Here, \(a=1, b=-6, c=k\), so \(D=b^2-4ac=(-6)^2-4(1)(k)=36-4k\). Setting \(D=0\) gives \(36-4k=0\), hence \(k=9\). The value 6 is only the magnitude of the coefficient of \(x\), so it is not correct. Exam tip: For equal roots of a quadratic equation, always use \(b^2-4ac=0\).
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Roots of a Quadratic Equation.
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