What condition on \(k\) is necessary for the equation \(x^2-6x+k=0\) to have two real and distinct roots?
Answer and explanation
Correct answer: \(k<9\)
For a quadratic equation \(ax^2+bx+c=0\) to have two real and distinct roots, its discriminant must satisfy \(\Delta=b^2-4ac>0\). Here, \(a=1, b=-6, c=k\), so \(\Delta=(-6)^2-4(1)(k)=36-4k\). Thus, \(36-4k>0\), which gives \(k<9\). When \(k=9\), the roots are equal, and when \(k>9\), the roots are not real. Exam tip: for distinct real roots, always require \(\Delta>0\).
Frequently asked questions
What is the correct answer to this question?
\(k<9\)
Why is this the correct answer?
For a quadratic equation \(ax^2+bx+c=0\) to have two real and distinct roots, its discriminant must satisfy \(\Delta=b^2-4ac>0\). Here, \(a=1, b=-6, c=k\), so \(\Delta=(-6)^2-4(1)(k)=36-4k\). Thus, \(36-4k>0\), which gives \(k<9\). When \(k=9\), the roots are equal, and when \(k>9\), the roots are not real. Exam tip: for distinct real roots, always require \(\Delta>0\).
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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