If \(x=\sqrt{2}+\sqrt{5}\), which polynomial equation is satisfied by \(x\)?
Answer and explanation
Correct answer: \(x^4-14x^2+9=0\)
Remove radicals by squaring. Compute \(x^2=(\sqrt{2}+\sqrt{5})^2=2+5+2\sqrt{10}=7+2\sqrt{10}\), so \(x^2-7=2\sqrt{10}\). Squaring again gives \((x^2-7)^2=40\), hence \(x^4-14x^2+49=40\) and therefore \(x^4-14x^2+9=0\). Thus option B is correct. Option C is wrong because it would force \(2\sqrt{10}=0\). Options A and D have incorrect signs/coefficients. Exam tip: for nested radicals square to eliminate them stepwise; using conjugates also helps find the minimal polynomial over Q.
Frequently asked questions
What is the correct answer to this question?
\(x^4-14x^2+9=0\)
Why is this the correct answer?
Remove radicals by squaring. Compute \(x^2=(\sqrt{2}+\sqrt{5})^2=2+5+2\sqrt{10}=7+2\sqrt{10}\), so \(x^2-7=2\sqrt{10}\). Squaring again gives \((x^2-7)^2=40\), hence \(x^4-14x^2+49=40\) and therefore \(x^4-14x^2+9=0\). Thus option B is correct. Option C is wrong because it would force \(2\sqrt{10}=0\). Options A and D have incorrect signs/coefficients. Exam tip: for nested radicals square to eliminate them stepwise; using conjugates also helps find the minimal polynomial over Q.
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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