If the equation \(x^2-2(a+2)x+(a^2+6a+8)=0\) has no real roots, which condition on \(a\) is correct?
Answer and explanation
Correct answer: \(a>-2\)
A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(D=[-2(a+2)]^2-4(a^2+6a+8)=-8(a+2)\). Thus, \(-8(a+2)<0\), which gives \(a>-2\). When \(a=-2\), \(D=0\), so the equation has two equal real roots; hence it is not correct. Exam tip: For ‘no real roots’, always apply the condition \(D<0\).
Frequently asked questions
What is the correct answer to this question?
\(a>-2\)
Why is this the correct answer?
A quadratic equation has no real roots when its discriminant satisfies \(D<0\). Here, \(D=[-2(a+2)]^2-4(a^2+6a+8)=-8(a+2)\). Thus, \(-8(a+2)<0\), which gives \(a>-2\). When \(a=-2\), \(D=0\), so the equation has two equal real roots; hence it is not correct. Exam tip: For ‘no real roots’, always apply 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.
Student feedback
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
