Which option gives the correct simplified form of (\sqrt{50}+3\sqrt{8}-\sqrt{18})?
(\sqrt{50}=5\sqrt{2}), (3\sqrt{8}=6\sqrt{2}) and (\sqrt{18}=3\sqrt{2}). Hence the value is (8\sqrt{2}).
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 8 questions from this page. Select your focus, then start.
(\sqrt{50}=5\sqrt{2}), (3\sqrt{8}=6\sqrt{2}) and (\sqrt{18}=3\sqrt{2}). Hence the value is (8\sqrt{2}).
(x^2=3+2\sqrt{2}) and (x^3=7+5\sqrt{2}), so (x^3-3x=4+2\sqrt{2}). The correct value is not in the options so calculate carefully.
(x^3=7+5\sqrt{2}) and (3x=3+3\sqrt{2}), so the difference is (4+2\sqrt{2}). In exams calculate powers step by step.
The sum of (9+\sqrt{7}) and (9-\sqrt{7}) is (18), and the product is (81-7=74). In exams check both sum and product of options.
Multiplying by the conjugate gives (\frac{30+2\sqrt{221}}{4}=\frac{15+\sqrt{221}}{2}). In exams divide by the common factor at the end.
After cancelling (385=5\cdot7\cdot11), only (2^3) remains in the denominator. In exams always check the denominator in lowest form.
The first product is (14-6=8), and (\sqrt{84}=2\sqrt{21}). In exams use both conjugate multiplication and radical simplification.
The other zero will be (6-\sqrt{19}), and the product is (36-19=17). In exams connect the constant term with the product of zeroes.
QUIZ COMPLETE