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Which of the following polynomials has zeros \(3+\sqrt{2}\) and \(3-\sqrt{2}\)?

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Answer and explanation

Correct answer: \(x^2-6x+7\)

For zeros \(3+\sqrt{2}\) and \(3-\sqrt{2}\) the sum is \((3+\sqrt{2})+(3-\sqrt{2})=6\) and the product is \((3+\sqrt{2})(3-\sqrt{2})=9-2=7\). A monic quadratic with these zeros is \(x^2-(\text{sum})x+(\text{product})=x^2-6x+7\). The closest distractor \(x^2-6x+11\) (option D) has the same linear coefficient but the constant term (product) is incorrect. Exam tip: form \((x-(3+\sqrt{2}))(x-(3-\sqrt{2}))\) and expand — this avoids sign mistakes.

Related tags

Polynomial-FormationConjugate-RootsQuadratic-EquationIrrational-Roots

Frequently asked questions

What is the correct answer to this question?

\(x^2-6x+7\)

Why is this the correct answer?

For zeros \(3+\sqrt{2}\) and \(3-\sqrt{2}\) the sum is \((3+\sqrt{2})+(3-\sqrt{2})=6\) and the product is \((3+\sqrt{2})(3-\sqrt{2})=9-2=7\). A monic quadratic with these zeros is \(x^2-(\text{sum})x+(\text{product})=x^2-6x+7\). The closest distractor \(x^2-6x+11\) (option D) has the same linear coefficient but the constant term (product) is incorrect. Exam tip: form \((x-(3+\sqrt{2}))(x-(3-\sqrt{2}))\) and expand — this avoids sign mistakes.

Which subject and chapter does this question cover?

This is a Class 10 Mathematics question. Chapter: Polynomials. Topic: Irrational numbers and real numbers.

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