Muft Shiksha™ एक 100% Free Education Portal है 🇮🇳, जिसका उद्देश्य Class 9–12 के हर विद्यार्थी तक High-Quality Education को पूरी तरह मुफ्त पहुँचाना है। 🇮🇳 हम मानते हैं कि अच्छी शिक्षा किसी student की आर्थिक स्थिति पर निर्भर नहीं होनी चाहिए। 🇮🇳 हर विद्यार्थी को वही Quality Study Material, MCQs, Quizzes, Exam Preparation, Concept-Based Learning और Bilingual Support मिलना चाहिए, जो आमतौर पर महंगी Coaching या Premium Platforms में मिलता है। Muft Shiksha™ 🇮🇳 इसी सोच के साथ बनाया गया है
SubjectsMathematics
अपरिमेय संख्याएँ और वास्तविक संख्याएँ
In this Class 10 Mathematics topic, students build a clear understanding of real numbers as the collection of rational and irrational numbers. They learn to identify irrational numbers, compare and represent real numbers on the number line, and interpret terminating, recurring, and non-terminating non-recurring decimals. The topic also develops confidence with properties and operations involving real numbers, providing useful foundations for reading polynomial expressions, coefficients, and real zeros in the surrounding Polynomials chapter.
TOPIC PRACTICE
Up to 3 questions from this page. Select your focus, then start.
For rational coefficients, irrational zeroes usually occur in conjugate pairs. Hence the companion zero of (3-\sqrt{5}) is (3+\sqrt{5}).
For the monic quadratic p(x) = x² − 2kx + 20, the product of the zeroes equals the constant term, 20. The stated zeroes have product (k + √5)(k − √5) = k² − 5 by the difference-of-squares identity. Equating these products gives k² − 5 = 20, so k² = 25 and k = ±5. Since the question asks for the positive value, k = 5, making option A correct. The sum provides an additional check: (k + √5) + (k − √5) = 2k, which agrees with the coefficient relation for x² − 2kx + 20. Options B, C, and √15 do not satisfy k² = 25 and therefore cannot produce the stated constant term.
The governing concept is rationalisation by multiplying by the conjugate. Since p=6−√35, its reciprocal is 1/(6−√35). Multiply numerator and denominator by 6+√35: 1/p=(6+√35)/[(6−√35)(6+√35)]=(6+√35)/(36−35)=6+√35. Now subtract p: 1/p−p=(6+√35)−(6−√35)=2√35. Therefore option A is correct. Option B incorrectly cancels the radical terms, option C loses the factor 2 produced when subtracting a negative radical, and option D introduces an unsupported factor 6. The denominator is non-zero because √35 is less than 6, so the reciprocal is well defined.
QUIZ COMPLETE
