For the equation \(x^2-(k+1)x+4=0\), in which interval must \(k\) lie for the equation to have no real roots?
Answer and explanation
Correct answer: \(-5<k<3\)
A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(a=1\), \(b=-(k+1)\), and \(c=4\), so \(D=(k+1)^2-16\). Thus, \((k+1)^2<16\), which gives \(-4<k+1<4\), and hence \(-5<k<3\). Therefore, option A is correct. At the boundary values \(k=-5\) and \(k=3\), \(D=0\), so the equation has two equal real roots rather than no real roots. Exam tip: for ‘no real roots’, always use the condition \(D<0\).
Frequently asked questions
What is the correct answer to this question?
\(-5<k<3\)
Why is this the correct answer?
A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(a=1\), \(b=-(k+1)\), and \(c=4\), so \(D=(k+1)^2-16\). Thus, \((k+1)^2<16\), which gives \(-4<k+1<4\), and hence \(-5<k<3\). Therefore, option A is correct. At the boundary values \(k=-5\) and \(k=3\), \(D=0\), so the equation has two equal real roots rather than no real roots. Exam tip: for ‘no real roots’, always use 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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