For real numbers a and b, if the equation \(x^2-2(a+b)x+(a-b)^2=0\) has real and distinct roots, which condition on a and b is correct?
Answer and explanation
Correct answer: \(ab>0\)
A quadratic equation has real and distinct roots only when its discriminant is positive. Here, \(A=1\), \(B=-2(a+b)\), and \(C=(a-b)^2\). Thus, \(D=[-2(a+b)]^2-4(a-b)^2=16ab\). The condition \(D>0\) gives \(16ab>0\), and hence \(ab>0\). If \(ab=0\), the roots are equal, while \(ab<0\) gives non-real roots. Exam tip: simplify the discriminant completely before applying its sign condition.
Frequently asked questions
What is the correct answer to this question?
\(ab>0\)
Why is this the correct answer?
A quadratic equation has real and distinct roots only when its discriminant is positive. Here, \(A=1\), \(B=-2(a+b)\), and \(C=(a-b)^2\). Thus, \(D=[-2(a+b)]^2-4(a-b)^2=16ab\). The condition \(D>0\) gives \(16ab>0\), and hence \(ab>0\). If \(ab=0\), the roots are equal, while \(ab<0\) gives non-real roots. Exam tip: simplify the discriminant completely before applying its sign condition.
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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