If \(p(x)=x^2+2\sqrt{3}x+3\), what are its zeros?
Answer and explanation
Correct answer: (-\sqrt{3}) twice
The quadratic is a perfect square: \(p(x)=(x+\sqrt{3})^2\). Expanding gives \((x+\sqrt{3})^2=x^2+2\sqrt{3}x+3\), so the only root is \(x=-\sqrt{3}\) with multiplicity two. Using the discriminant: \(\Delta=(2\sqrt{3})^2-4\cdot1\cdot3=12-12=0\), confirming a repeated real root. The closest distractor \(\sqrt{3}\) is wrong due to the sign. Exam tip: if \(\Delta=0\) or you can complete the square, expect a double root and compute \(-b/2a\).
Frequently asked questions
What is the correct answer to this question?
(-\sqrt{3}) twice
Why is this the correct answer?
The quadratic is a perfect square: \(p(x)=(x+\sqrt{3})^2\). Expanding gives \((x+\sqrt{3})^2=x^2+2\sqrt{3}x+3\), so the only root is \(x=-\sqrt{3}\) with multiplicity two. Using the discriminant: \(\Delta=(2\sqrt{3})^2-4\cdot1\cdot3=12-12=0\), confirming a repeated real root. The closest distractor \(\sqrt{3}\) is wrong due to the sign. Exam tip: if \(\Delta=0\) or you can complete the square, expect a double root and compute \(-b/2a\).
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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