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If \(p(x)=x^2+kx+9\) is a perfect-square polynomial and \(k>0\), what is the value of \(k\)?
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Answer and explanation
Correct answer: 6
Since the constant term is \(9=3^2\), the perfect-square form must be \((x+3)^2=x^2+6x+9\). Therefore, \(k=6\). Although \((x-3)^2\) gives \(k=-6\), it is excluded because \(k>0\). In exams, identify the middle term of a square as \(2ab\).
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What is the correct answer to this question?
6
Why is this the correct answer?
Since the constant term is \(9=3^2\), the perfect-square form must be \((x+3)^2=x^2+6x+9\). Therefore, \(k=6\). Although \((x-3)^2\) gives \(k=-6\), it is excluded because \(k>0\). In exams, identify the middle term of a square as \(2ab\).
Which subject and chapter does this question cover?
This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Polynomials in one variable.
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