\( \sqrt{108}=6\sqrt{3} \) and \( \sqrt{48}=4\sqrt{3} \), so the difference is \(2\sqrt{3}\). Subtract only like radicals.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{3}\). \( \sqrt{108}=6\sqrt{3} \) and \( \sqrt{48}=4\sqrt{3} \), so the difference is \(2\sqrt{3}\). Subtract only like radicals.
Step 3
Exam Tip
\( \sqrt{108}=6\sqrt{3} \) और \( \sqrt{48}=4\sqrt{3} \), इसलिए अंतर \(2\sqrt{3}\) है। समान मूलों को ही घटाएँ।
\( \sqrt{75}=5\sqrt{3} \) and \( \sqrt{27}=3\sqrt{3} \), so the difference is \(2\sqrt{3}\). Subtract only like radicals.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{3}\). \( \sqrt{75}=5\sqrt{3} \) and \( \sqrt{27}=3\sqrt{3} \), so the difference is \(2\sqrt{3}\). Subtract only like radicals.
Step 3
Exam Tip
\( \sqrt{75}=5\sqrt{3} \) और \( \sqrt{27}=3\sqrt{3} \), इसलिए अंतर \(2\sqrt{3}\) है। समान मूलों को ही घटाएँ।
\( \sqrt{27}=3\sqrt{3} \) and \( \sqrt{12}=2\sqrt{3} \), so the difference is \( \sqrt{3} \). Subtract like radicals.
Step 2
Why this answer is correct
The correct answer is A. \( \sqrt{3} \). \( \sqrt{27}=3\sqrt{3} \) and \( \sqrt{12}=2\sqrt{3} \), so the difference is \( \sqrt{3} \). Subtract like radicals.
Step 3
Exam Tip
\( \sqrt{27}=3\sqrt{3} \) और \( \sqrt{12}=2\sqrt{3} \) इसलिए अंतर \( \sqrt{3} \) है। समान मूलों को घटाएँ।
A. यह (0) और (1) के बीच अपरिमेय है/It is irrational between (0) and (1)
Step 1
Concept
This decimal is non-terminating and non-repeating, so it is irrational and lies between (0) and (1). Identify rationality by the decimal pattern.
Step 2
Why this answer is correct
The correct answer is A. यह (0) और (1) के बीच अपरिमेय है / It is irrational between (0) and (1). This decimal is non-terminating and non-repeating, so it is irrational and lies between (0) and (1). Identify rationality by the decimal pattern.
Step 3
Exam Tip
यह दशमलव अनावर्ती और असांत है, इसलिए अपरिमेय है और (0) से (1) के बीच है। दशमलव पैटर्न देखकर परिमेयता पहचानें।
\(\sqrt{3}\approx1.732\), so it lies near (1.732) on the number line. Remember approximations of common square roots.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{3}\). \(\sqrt{3}\approx1.732\), so it lies near (1.732) on the number line. Remember approximations of common square roots.
Step 3
Exam Tip
\(\sqrt{3}\approx1.732\), इसलिए संख्या रेखा पर (1.732) के पास यही होगा। कुछ प्रसिद्ध वर्गमूलों के अनुमान याद रखें।
Since (17=\(\sqrt{17}\)2) and the middle term is \(-2\sqrt{17}x\), it is (\(x-\sqrt{17}\)2). In exams, identify perfect squares even with irrational coefficients.
Step 2
Why this answer is correct
The correct answer is A. (\(x-\sqrt{17}\)2=0). Since (17=\(\sqrt{17}\)2) and the middle term is \(-2\sqrt{17}x\), it is (\(x-\sqrt{17}\)2). In exams, identify perfect squares even with irrational coefficients.
Step 3
Exam Tip
क्योंकि (17=\(\sqrt{17}\)2) और मध्य पद \(-2\sqrt{17}x\) है, इसलिए यह (\(x-\sqrt{17}\)2) है। परीक्षा में अपरिमेय गुणांक में भी पूर्ण वर्ग पहचानें।
Since (13=\(\sqrt{13}\)2) and the middle term is \(2\sqrt{13}x\), it is (\(x+\sqrt{13}\)2). In exams, identify perfect squares even with irrational coefficients.
Step 2
Why this answer is correct
The correct answer is A. (\(x+\sqrt{13}\)2=0). Since (13=\(\sqrt{13}\)2) and the middle term is \(2\sqrt{13}x\), it is (\(x+\sqrt{13}\)2). In exams, identify perfect squares even with irrational coefficients.
Step 3
Exam Tip
क्योंकि (13=\(\sqrt{13}\)2) और मध्य पद \(2\sqrt{13}x\) है, इसलिए यह (\(x+\sqrt{13}\)2) है। परीक्षा में अपरिमेय गुणांक में भी पूर्ण वर्ग पहचानें।
Since (11=\(\sqrt{11}\)2) and the middle term is \(-2\sqrt{11}x\), it is (\(x-\sqrt{11}\)2). In exams, identify perfect squares even with irrational coefficients.
Step 2
Why this answer is correct
The correct answer is A. (\(x-\sqrt{11}\)2=0). Since (11=\(\sqrt{11}\)2) and the middle term is \(-2\sqrt{11}x\), it is (\(x-\sqrt{11}\)2). In exams, identify perfect squares even with irrational coefficients.
Step 3
Exam Tip
क्योंकि (11=\(\sqrt{11}\)2) और मध्य पद \(-2\sqrt{11}x\) है, इसलिए यह (\(x-\sqrt{11}\)2) है। परीक्षा में अपरिमेय गुणांक में भी पूर्ण वर्ग पहचानें।
Since (7=\(\sqrt{7}\)2) and the middle term is \(2\sqrt{7}x\), it is (\(x+\sqrt{7}\)2). In exams, identify perfect squares even with irrational coefficients.
Step 2
Why this answer is correct
The correct answer is A. (\(x+\sqrt{7}\)2=0). Since (7=\(\sqrt{7}\)2) and the middle term is \(2\sqrt{7}x\), it is (\(x+\sqrt{7}\)2). In exams, identify perfect squares even with irrational coefficients.
Step 3
Exam Tip
क्योंकि (7=\(\sqrt{7}\)2) और मध्य पद \(2\sqrt{7}x\) है, इसलिए यह (\(x+\sqrt{7}\)2) है। परीक्षा में अपरिमेय गुणांक में भी पूर्ण वर्ग पहचानें।
Since (5=\(\sqrt{5}\)2) and the middle term is \(-2\sqrt{5}x\), it is (\(x-\sqrt{5}\)2). In exams, identify perfect squares even with irrational coefficients.
Step 2
Why this answer is correct
The correct answer is A. (\(x-\sqrt{5}\)2=0). Since (5=\(\sqrt{5}\)2) and the middle term is \(-2\sqrt{5}x\), it is (\(x-\sqrt{5}\)2). In exams, identify perfect squares even with irrational coefficients.
Step 3
Exam Tip
क्योंकि (5=\(\sqrt{5}\)2) और मध्य पद \(-2\sqrt{5}x\) है, इसलिए यह (\(x-\sqrt{5}\)2) है। परीक्षा में अपरिमेय गुणांक में भी पूर्ण वर्ग पहचानें।
Since (3=\(\sqrt{3}\)2) and the middle term is \(2\sqrt{3}x\), it is (\(x+\sqrt{3}\)2). In exams, identify perfect squares even with irrational coefficients.
Step 2
Why this answer is correct
The correct answer is A. (\(x+\sqrt{3}\)2=0). Since (3=\(\sqrt{3}\)2) and the middle term is \(2\sqrt{3}x\), it is (\(x+\sqrt{3}\)2). In exams, identify perfect squares even with irrational coefficients.
Step 3
Exam Tip
क्योंकि (3=\(\sqrt{3}\)2) और मध्य पद \(2\sqrt{3}x\) है, इसलिए यह (\(x+\sqrt{3}\)2) है। परीक्षा में अपरिमेय गुणांक में भी पूर्ण वर्ग पहचानें।
\(\sqrt{45}=3\sqrt{5}\), which is real and irrational. In exams do not treat the square root of a negative number as real.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{45}\). \(\sqrt{45}=3\sqrt{5}\), which is real and irrational. In exams do not treat the square root of a negative number as real.
Step 3
Exam Tip
\(\sqrt{45}=3\sqrt{5}\), जो वास्तविक और अपरिमेय है। परीक्षा में ऋणात्मक वर्गमूल को वास्तविक संख्या न मानें।
\(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\), so the difference is \(\sqrt{3}\). Simplify first in exams.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{3}\). \(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\), so the difference is \(\sqrt{3}\). Simplify first in exams.
Step 3
Exam Tip
\(\sqrt{27}=3\sqrt{3}\) और \(\sqrt{12}=2\sqrt{3}\), इसलिए अंतर \(\sqrt{3}\) है। परीक्षा में पहले सरलीकरण करें।
A. \(3\sqrt{2}\), अपरिमेय/\(3\sqrt{2}\), irrational
Step 1
Concept
\(\sqrt{8}=2\sqrt{2}\), so \(a=3\sqrt{2}\), irrational. Combine like radicals in exams.
Step 2
Why this answer is correct
The correct answer is A. \(3\sqrt{2}\), अपरिमेय / \(3\sqrt{2}\), irrational. \(\sqrt{8}=2\sqrt{2}\), so \(a=3\sqrt{2}\), irrational. Combine like radicals in exams.
Step 3
Exam Tip
\(\sqrt{8}=2\sqrt{2}\), इसलिए \(a=3\sqrt{2}\) अपरिमेय है। परीक्षा में समान करणी वाले पद जोड़ें।
(p(x)=x\(x-\sqrt{5}\)), so the zeroes are (0) and \(\sqrt{5}\). Taking the common factor is a fast method in exams.
Step 2
Why this answer is correct
The correct answer is A. \(0,\sqrt{5}\). (p(x)=x\(x-\sqrt{5}\)), so the zeroes are (0) and \(\sqrt{5}\). Taking the common factor is a fast method in exams.
Step 3
Exam Tip
(p(x)=x\(x-\sqrt{5}\)), इसलिए शून्यक (0) और \(\sqrt{5}\) हैं। परीक्षा में सामान्य गुणनखंड निकालना तेज तरीका है।
Because (\(3+\sqrt{8}\)\(3-\sqrt{8}\)=9-8=1), the reciprocal is \(3-\sqrt{8}\). If the product is (1), the reciprocal is immediate.
Step 2
Why this answer is correct
The correct answer is A. \(3-\sqrt{8}\). Because (\(3+\sqrt{8}\)\(3-\sqrt{8}\)=9-8=1), the reciprocal is \(3-\sqrt{8}\). If the product is (1), the reciprocal is immediate.
Step 3
Exam Tip
क्योंकि (\(3+\sqrt{8}\)\(3-\sqrt{8}\)=9-8=1), इसलिए व्युत्क्रम \(3-\sqrt{8}\) है। परीक्षा में गुणनफल (1) होने पर व्युत्क्रम तुरंत मिल जाता है।
A. शून्यकों का गुणनफल \(-3\sqrt{2}\) है/The product of zeroes is \(-3\sqrt{2}\)
Step 1
Concept
In a monic quadratic, the constant term is the product of zeroes. Here \(\alpha\beta=-3\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. शून्यकों का गुणनफल \(-3\sqrt{2}\) है / The product of zeroes is \(-3\sqrt{2}\). In a monic quadratic, the constant term is the product of zeroes. Here \(\alpha\beta=-3\sqrt{2}\).
Step 3
Exam Tip
एकक द्विघात में स्थिर पद शून्यकों का गुणनफल होता है। यहाँ \(\alpha\beta=-3\sqrt{2}\) है।
The sum is \(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\). Simplify radicals before giving the final answer.
Step 2
Why this answer is correct
The correct answer is A. \(3\sqrt{2}\). The sum is \(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\). Simplify radicals before giving the final answer.
Step 3
Exam Tip
योग \(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\) है। मूलों को सरल करके ही अंतिम उत्तर दें।
A. दूसरा \(\sqrt{5}\), \(k=\sqrt{5}\)/Other \(\sqrt{5}\), \(k=\sqrt{5}\)
Step 1
Concept
The product is (5), so the other zero is \(\frac{5}{\sqrt{5}}=\sqrt{5}\). The sum is \(2\sqrt{5}=2k\), hence \(k=\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is A. दूसरा \(\sqrt{5}\), \(k=\sqrt{5}\) / Other \(\sqrt{5}\), \(k=\sqrt{5}\). The product is (5), so the other zero is \(\frac{5}{\sqrt{5}}=\sqrt{5}\). The sum is \(2\sqrt{5}=2k\), hence \(k=\sqrt{5}\).
Step 3
Exam Tip
गुणनफल (5) है, इसलिए दूसरा शून्यक \(\frac{5}{\sqrt{5}}=\sqrt{5}\) होगा। योग \(2\sqrt{5}=2k\), अतः \(k=\sqrt{5}\) है।
