The equation \(x^2+14x+k=0\) has no real roots. What is the correct condition on \(k\)?
Answer and explanation
Correct answer: \(k>49\)
For no real roots the discriminant \(D=b^2-4ac\) must be negative. Here \(a=1,\; b=14,\; c=k\), so \(D=14^2-4\cdot1\cdot k=196-4k\). Requiring \(D<0\) gives \(196-4k<0\Rightarrow k>49\). Thus \(k>49\) is correct. The closest distractor \(k=49\) yields \(D=0\), giving one repeated real root, not “no real roots.” Exam tip: compute \(D\) first and use its sign to decide the nature of roots (\(D<0,=0,>0\)).
Frequently asked questions
What is the correct answer to this question?
\(k>49\)
Why is this the correct answer?
For no real roots the discriminant \(D=b^2-4ac\) must be negative. Here \(a=1,\; b=14,\; c=k\), so \(D=14^2-4\cdot1\cdot k=196-4k\). Requiring \(D<0\) gives \(196-4k<0\Rightarrow k>49\). Thus \(k>49\) is correct. The closest distractor \(k=49\) yields \(D=0\), giving one repeated real root, not “no real roots.” Exam tip: compute \(D\) first and use its sign to decide the nature of roots (\(D<0,=0,>0\)).
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Introduction to Quadratic Equations.
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