For the equation \(x^2-2(k+1)x+k^2=0\) to have real and distinct roots, what is the correct condition on \(k\)?
Answer and explanation
Correct answer: \(k>- rac{1}{2}\)
A quadratic equation has real and distinct roots only when its discriminant satisfies \(D>0\). Here, \(a=1\), \(b=-2(k+1)\), and \(c=k^2\), so \(D=b^2-4ac=4(k+1)^2-4k^2=4(2k+1)\). Thus, \(4(2k+1)>0\), which gives \(k>- rac{1}{2}\). Option B is incorrect because at \(k=- rac{1}{2}\), the roots are equal rather than distinct. Exam tip: For ‘real and distinct’ roots, use the strict condition \(D>0\).
Frequently asked questions
What is the correct answer to this question?
\(k>- rac{1}{2}\)
Why is this the correct answer?
A quadratic equation has real and distinct roots only when its discriminant satisfies \(D>0\). Here, \(a=1\), \(b=-2(k+1)\), and \(c=k^2\), so \(D=b^2-4ac=4(k+1)^2-4k^2=4(2k+1)\). Thus, \(4(2k+1)>0\), which gives \(k>- rac{1}{2}\). Option B is incorrect because at \(k=- rac{1}{2}\), the roots are equal rather than distinct. Exam tip: For ‘real and distinct’ roots, use the strict condition \(D>0\).
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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