If the roots of the quadratic equation \(x^2-(m+1)x+m=0\) are reciprocals of each other, what is the value of \(m\)?
Answer and explanation
Correct answer: 1
Let the roots be \(\alpha\) and \(\beta\). By Vieta’s formula, \(\alpha\beta=m\). For two roots to be reciprocals of each other, \(\alpha\beta=1\); hence, \(m=1\). Indeed, when \(m=1\), the equation becomes \((x-1)^2=0\), so both roots are 1, and each is the reciprocal of the other. The value \(m=0\) is invalid because it gives a zero root, whose reciprocal is undefined. Exam tip: For reciprocal roots, set their product equal to 1.
Frequently asked questions
What is the correct answer to this question?
1
Why is this the correct answer?
Let the roots be \(\alpha\) and \(\beta\). By Vieta’s formula, \(\alpha\beta=m\). For two roots to be reciprocals of each other, \(\alpha\beta=1\); hence, \(m=1\). Indeed, when \(m=1\), the equation becomes \((x-1)^2=0\), so both roots are 1, and each is the reciprocal of the other. The value \(m=0\) is invalid because it gives a zero root, whose reciprocal is undefined. Exam tip: For reciprocal roots, set their product equal to 1.
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Quadratic Equations. Topic: Roots of a Quadratic Equation.
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