If \(p(x)=x^2-2\sqrt{2}x+1\), what is the type of its zeroes?
Answer and explanation
Correct answer: Two distinct real irrational
Use the discriminant \(D=b^2-4ac\). Here \(a=1,\;b=-2\sqrt{2},\;c=1\), so \(D=( -2\sqrt{2})^2-4\cdot1\cdot1=8-4=4>0\). Thus the roots are real and distinct. Applying the quadratic formula gives roots \(\frac{-b\pm\sqrt{D}}{2a}=\frac{2\sqrt{2}\pm2}{2}=\sqrt{2}\pm1\). Since \(\sqrt{2}\) is irrational, \(\sqrt{2}\pm1\) are also irrational. Therefore the zeros are two distinct real irrational numbers. The closest distractor (two equal roots) is incorrect because \(D\neq0\). Exam tip: check \(D\) first to determine reality and multiplicity of roots, then inspect any irrational parts (like \(\sqrt{2}\)) to decide rationality.
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What is the correct answer to this question?
Two distinct real irrational
Why is this the correct answer?
Use the discriminant \(D=b^2-4ac\). Here \(a=1,\;b=-2\sqrt{2},\;c=1\), so \(D=( -2\sqrt{2})^2-4\cdot1\cdot1=8-4=4>0\). Thus the roots are real and distinct. Applying the quadratic formula gives roots \(\frac{-b\pm\sqrt{D}}{2a}=\frac{2\sqrt{2}\pm2}{2}=\sqrt{2}\pm1\). Since \(\sqrt{2}\) is irrational, \(\sqrt{2}\pm1\) are also irrational. Therefore the zeros are two distinct real irrational numbers. The closest distractor (two equal roots) is incorrect because \(D\neq0\). Exam tip: check \(D\) first to determine reality and multiplicity of roots, then inspect any irrational parts (like \(\sqrt{2}\)) to decide rationality.
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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