If \(x\) is a real number, what is the minimum value of \(p(x)=x^2+10x+29\)?
Answer and explanation
Correct answer: 4
Completing the square gives \(p(x)=x^2+10x+29=(x+5)^2+4\). Since \((x+5)^2\geq 0\), we have \(p(x)\geq 4\), and equality occurs at \(x=-5\). Therefore, the minimum value is 4. The value 29 is only the constant term, not the minimum. Exam tip: Rewrite a quadratic in the form \((x-a)^2+k\) to identify its minimum quickly.
Frequently asked questions
What is the correct answer to this question?
4
Why is this the correct answer?
Completing the square gives \(p(x)=x^2+10x+29=(x+5)^2+4\). Since \((x+5)^2\geq 0\), we have \(p(x)\geq 4\), and equality occurs at \(x=-5\). Therefore, the minimum value is 4. The value 29 is only the constant term, not the minimum. Exam tip: Rewrite a quadratic in the form \((x-a)^2+k\) to identify its minimum quickly.
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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