For the quadratic equation \(3x^2-2(2k+1)x+(k+1)^2=0\) to have equal roots, what are the values of \(k\)?
Answer and explanation
Correct answer: \(k=1+\sqrt{3}\) or \(k=1-\sqrt{3}\)
A quadratic equation has equal roots when its discriminant \(D=b^2-4ac\) is zero. Here, \(a=3\), \(b=-2(2k+1)\), and \(c=(k+1)^2\). Thus, \(D=4(2k+1)^2-12(k+1)^2=4(k^2-2k-2)\). Setting this equal to zero gives \(k^2-2k-2=0\), so \(k=1\pm\sqrt{3}\). The values in option B result from incorrect factorisation. Exam tip: For equal roots, immediately apply the condition \(D=0\).
Frequently asked questions
What is the correct answer to this question?
\(k=1+\sqrt{3}\) or \(k=1-\sqrt{3}\)
Why is this the correct answer?
A quadratic equation has equal roots when its discriminant \(D=b^2-4ac\) is zero. Here, \(a=3\), \(b=-2(2k+1)\), and \(c=(k+1)^2\). Thus, \(D=4(2k+1)^2-12(k+1)^2=4(k^2-2k-2)\). Setting this equal to zero gives \(k^2-2k-2=0\), so \(k=1\pm\sqrt{3}\). The values in option B result from incorrect factorisation. Exam tip: For equal roots, immediately apply the condition \(D=0\).
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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