For the equation \((k-2)x^2-2(k+1)x+k=0\) to have real and equal roots, what should be the value of \(k\)?
Answer and explanation
Correct answer: \(k=-\frac{1}{4}\)
Here, \(a=k-2\), \(b=-2(k+1)\), and \(c=k\). For real and equal roots, the discriminant must be zero: \(D=b^2-4ac=0\). Thus, \(D=4(k+1)^2-4(k-2)k=16k+4\). Setting this equal to zero gives \(k=-\frac{1}{4}\). For this value, \(a\neq 0\), so the equation remains quadratic. Exam tip: For equal-root questions, set the discriminant to zero and then verify that the coefficient of \(x^2\) is non-zero.
Frequently asked questions
What is the correct answer to this question?
\(k=-\frac{1}{4}\)
Why is this the correct answer?
Here, \(a=k-2\), \(b=-2(k+1)\), and \(c=k\). For real and equal roots, the discriminant must be zero: \(D=b^2-4ac=0\). Thus, \(D=4(k+1)^2-4(k-2)k=16k+4\). Setting this equal to zero gives \(k=-\frac{1}{4}\). For this value, \(a\neq 0\), so the equation remains quadratic. Exam tip: For equal-root questions, set the discriminant to zero and then verify that the coefficient of \(x^2\) is non-zero.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Nature of Roots.
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