If the equation \(kx^2-2(k+1)x+(k+4)=0\) has equal roots, what is the value of \(k\)?
Answer and explanation
Correct answer: \(\frac{1}{2}\)
For a quadratic equation \(ax^2+bx+c=0\) to have equal roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=k\), \(b=-2(k+1)\), and \(c=k+4\). Therefore, \(D=4(k+1)^2-4k(k+4)=4(1-2k)\). Setting this equal to zero gives \(k=\frac{1}{2}\). Exam tip: In equal-root questions, begin with \(D=0\) and verify that \(a\neq 0\) so the equation remains quadratic.
Frequently asked questions
What is the correct answer to this question?
\(\frac{1}{2}\)
Why is this the correct answer?
For a quadratic equation \(ax^2+bx+c=0\) to have equal roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=k\), \(b=-2(k+1)\), and \(c=k+4\). Therefore, \(D=4(k+1)^2-4k(k+4)=4(1-2k)\). Setting this equal to zero gives \(k=\frac{1}{2}\). Exam tip: In equal-root questions, begin with \(D=0\) and verify that \(a\neq 0\) so the equation remains quadratic.
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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