If a and b are real numbers, which is the correct condition for the equation \(x^2-2(a-b)x+(a+b)^2=0\) to have real roots?
Answer and explanation
Correct answer: ab ≤ 0
The discriminant of the quadratic equation is \(D=[-2(a-b)]^2-4(a+b)^2\). Therefore, \(D=4[(a-b)^2-(a+b)^2]=-16ab\). For real roots, \(D\geq0\), so \(-16ab\geq0\), which gives \(ab\leq0\). Option B reverses the inequality. The condition \(a+b=0\) may satisfy the requirement, but it is not the complete necessary condition. Exam tip: begin with \(D\geq0\) whenever the nature of roots is asked.
Frequently asked questions
What is the correct answer to this question?
ab ≤ 0
Why is this the correct answer?
The discriminant of the quadratic equation is \(D=[-2(a-b)]^2-4(a+b)^2\). Therefore, \(D=4[(a-b)^2-(a+b)^2]=-16ab\). For real roots, \(D\geq0\), so \(-16ab\geq0\), which gives \(ab\leq0\). Option B reverses the inequality. The condition \(a+b=0\) may satisfy the requirement, but it is not the complete necessary condition. Exam tip: begin with \(D\geq0\) whenever the nature of roots is asked.
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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