100 results found for "equivalent-radicals" in Class 10.
Which equation is equivalent to \(x^2-6x+9=0\)?
A. ((x-3)2=0)
((x-3)2=x-2-6x+9). Recognising perfect squares saves time.
The correct answer is A. ((x-3)2=0). ((x-3)2=x-2-6x+9). Recognising perfect squares saves time.
((x-3)2=x-2-6x+9) होता है। पूर्ण वर्ग पहचानने से समय बचता है।
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Which equation is equivalent to \(x^2+2x+1=0\)?
A. \((x+1)^2=0\)
((x+1)2=x-2+2x+1). Recognising perfect squares helps solve faster.
The correct answer is A. \((x+1)^2=0\). ((x+1)2=x-2+2x+1). Recognising perfect squares helps solve faster.
((x+1)2=x-2+2x+1) होता है। पूर्ण वर्ग पहचानना तेजी से हल कराने में मदद करता है।
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Which number is equal to \(7\sqrt{2}\)?
A. \(\sqrt{98}\)
\(7\sqrt{2}=\sqrt{49}\sqrt{2}\).
This equals \(\sqrt{98}\).
To move the outside coefficient inside, multiply by its square. चरण 1: \(7\sqrt{2}=\sqrt{49}\sqrt{2}\)। चरण 2: यह \(\sqrt{98}\) के बराबर है। चरण 3: बाहर का गुणांक अंदर ले जाने के लिए उसका वर्ग अंदर गुणा करें।
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Which number is equal to \(5\sqrt{3}\)?
A. \(\sqrt{75}\)
\(5\sqrt{3}=\sqrt{25}\sqrt{3}\).
This equals \(\sqrt{75}\).
When moving an outside coefficient inside the root, multiply by its square. चरण 1: \(5\sqrt{3}=\sqrt{25}\sqrt{3}\)। चरण 2: यह \(\sqrt{75}\) के बराबर है। चरण 3: बाहर के गुणांक को अंदर ले जाते समय उसका वर्ग अंदर गुणा करें।
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Which number is equal to \(6\sqrt{2}\)?
A. \(\sqrt{72}\)
\(6\sqrt{2}=\sqrt{36}\sqrt{2}\).
This equals \(\sqrt{72}\).
To move the outside coefficient inside, multiply by its square. चरण 1: \(6\sqrt{2}=\sqrt{36}\sqrt{2}\)। चरण 2: यह \(\sqrt{72}\) के बराबर है। चरण 3: बाहर का गुणांक अंदर ले जाने के लिए उसका वर्ग अंदर गुणा करें।
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Which number is equal to \(4\sqrt{3}\)?
A. \(\sqrt{48}\)
\(4\sqrt{3}=\sqrt{16}\sqrt{3}\).
This equals \(\sqrt{48}\).
When moving an outside coefficient inside the root, multiply by its square. चरण 1: \(4\sqrt{3}=\sqrt{16}\sqrt{3}\)। चरण 2: यह \(\sqrt{48}\) के बराबर है। चरण 3: बाहर के गुणांक को अंदर ले जाते समय उसका वर्ग गुणा करें।
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Which number is equal to \(3\sqrt{5}\)?
A. \(\sqrt{45}\)
\(3\sqrt{5}=\sqrt{9}\sqrt{5}\).
This equals \(\sqrt{45}\).
An outside coefficient can be moved inside by squaring it. चरण 1: \(3\sqrt{5}=\sqrt{9}\sqrt{5}\)। चरण 2: यह \(\sqrt{45}\) के बराबर है। चरण 3: बाहर के गुणांक को वर्ग करके अंदर ले जा सकते हैं।
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Which number is equal to \(\sqrt{50}\)?
A. \(5\sqrt{2}\)
Write \(50=25\times2\).
\(\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}\).
To identify an equivalent form, simplify the square root first. चरण 1: \(50=25\times2\) लिखें। चरण 2: \(\sqrt{50}=\sqrt{25\times2}=5\sqrt{2}\)। चरण 3: बराबर रूप पहचानने के लिए सबसे पहले वर्गमूल को सरल करें।
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Which number is equal to \(\sqrt{75}\)?
B. \(5\sqrt{3}\)
\(75=25 \times 3\).
\(\sqrt{75}=5\sqrt{3}\).
To identify an equivalent form, simplify the square root first. चरण 1: \(75=25 \times 3\) है। चरण 2: \(\sqrt{75}=5\sqrt{3}\)। चरण 3: बराबर रूप पहचानने के लिए पहले वर्गमूल को सरल करें।
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Which number is equal to \(\sqrt{12}\)?
A. \(2\sqrt{3}\)
\(12=4 \times 3\).
\(\sqrt{12}=2\sqrt{3}\).
To identify an equivalent form, simplify the square root first. चरण 1: \(12=4 \times 3\) है। चरण 2: \(\sqrt{12}=2\sqrt{3}\)। चरण 3: बराबर रूप पहचानने के लिए वर्गमूल को सरल करना जरूरी है।
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Which statement is correct about the equations (6x+3y=39) and (2x+y=13)?
D. पहला समीकरण दूसरे का (3) गुना हैThe first equation is (3) times the second
Multiplying (2x+y=13) by (3) gives (6x+3y=39). Recognizing equivalent equations is also useful in exams.
The correct answer is D. पहला समीकरण दूसरे का (3) गुना है / The first equation is (3) times the second. Multiplying (2x+y=13) by (3) gives (6x+3y=39). Recognizing equivalent equations is also useful in exams.
(2x+y=13) को (3) से गुणा करने पर (6x+3y=39) मिलता है। समान समीकरणों को पहचानना भी परीक्षा में उपयोगी है।
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If (6x+2y=28) and (3x+y=14), which statement is correct?
A. पहला समीकरण दूसरे का (2) गुना हैThe first equation is (2) times the second
Multiplying (3x+y=14) by (2) gives (6x+2y=28). Recognize equivalent equations formed by multiplication.
The correct answer is A. पहला समीकरण दूसरे का (2) गुना है / The first equation is (2) times the second. Multiplying (3x+y=14) by (2) gives (6x+2y=28). Recognize equivalent equations formed by multiplication.
(3x+y=14) को (2) से गुणा करने पर (6x+2y=28) मिलता है। गुणन से बने समान समीकरणों को पहचानें।
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If (4x+ay=20) is coincident with the line (2x+3y=10), what will be the value of (a)?
A. (6)
Multiplying (2x+3y=10) by (2) gives (4x+6y=20). Therefore (a=6).
The correct answer is A. (6). Multiplying (2x+3y=10) by (2) gives (4x+6y=20). Therefore (a=6).
(2x+3y=10) को (2) से गुणा करने पर (4x+6y=20) मिलता है। इसलिए (a=6) होगा।
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Which equation will give a coincident line with (2x+5y=13)?
A. (4x+10y=26)
Dividing (4x+10y=26) by (2) gives (2x+5y=13). Therefore both are the same line on the graph.
The correct answer is A. (4x+10y=26). Dividing (4x+10y=26) by (2) gives (2x+5y=13). Therefore both are the same line on the graph.
(4x+10y=26) को (2) से भाग देने पर (2x+5y=13) मिलता है। इसलिए दोनों ग्राफ पर एक ही रेखा हैं।
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Which equation will give a coincident line with (x+2y=9)?
A. (2x+4y=18)
Dividing (2x+4y=18) by (2) gives (x+2y=9). Therefore both are the same line on the graph.
The correct answer is A. (2x+4y=18). Dividing (2x+4y=18) by (2) gives (x+2y=9). Therefore both are the same line on the graph.
(2x+4y=18) को (2) से भाग देने पर (x+2y=9) मिलता है। इसलिए दोनों ग्राफ पर एक ही रेखा हैं।
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If (A=-8.375), \(B=-\sqrt{70}\), and \(C=-\frac{67}{8}\), which two points are equal?
C. (A) और (C)(A) and (C)
\( -\frac{67}{8}=-8.375 \), so (A=C). Convert the fraction to a decimal for comparison.
The correct answer is C. (A) और (C) / (A) and (C). \( -\frac{67}{8}=-8.375 \), so (A=C). Convert the fraction to a decimal for comparison.
\( -\frac{67}{8}=-8.375 \), इसलिए (A=C) है। भिन्न को दशमलव में बदलकर तुलना करें।
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If (A=-7.125), \(B=-\sqrt{51}\), and \(C=-\frac{57}{8}\), which two points are equal?
C. (A) और (C)(A) and (C)
\( -\frac{57}{8}=-7.125 \), so (A=C). Convert the fraction to a decimal for comparison.
The correct answer is C. (A) और (C) / (A) and (C). \( -\frac{57}{8}=-7.125 \), so (A=C). Convert the fraction to a decimal for comparison.
\( -\frac{57}{8}=-7.125 \), इसलिए (A=C) है। भिन्न को दशमलव में बदलकर तुलना करें।
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A. (A) और (C)(A) and (C)
\( -\frac{25}{4}=-6.25 \), so (A=C). Convert the fraction to a decimal for comparison.
The correct answer is A. (A) और (C) / (A) and (C). \( -\frac{25}{4}=-6.25 \), so (A=C). Convert the fraction to a decimal for comparison.
\( -\frac{25}{4}=-6.25 \), इसलिए (A=C) है। भिन्न को दशमलव में बदलकर तुलना करें।
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On the number line, (0.125) is the same point as which fraction?
B. \(\frac{1}{8}\)
\(0.125=\frac{125}{1000}=\frac{1}{8}\). Start with denominator (10), (100), or (1000), then simplify.
The correct answer is B. \(\frac{1}{8}\). \(0.125=\frac{125}{1000}=\frac{1}{8}\). Start with denominator (10), (100), or (1000), then simplify.
\(0.125=\frac{125}{1000}=\frac{1}{8}\) है। दशमलव को हर (10), (100), या (1000) से शुरू करके सरल करें।
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Which fraction shows the same point as (-1.25) on the number line?
B. \(-\frac{5}{4}\)
\(-1.25=-\frac{125}{100}=-\frac{5}{4}\). Convert the decimal into a simplified fraction.
The correct answer is B. \(-\frac{5}{4}\). \(-1.25=-\frac{125}{100}=-\frac{5}{4}\). Convert the decimal into a simplified fraction.
\(-1.25=-\frac{125}{100}=-\frac{5}{4}\) है। दशमलव को सरल भिन्न में बदलें।
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Which fraction represents the same point as (2.125) on the number line?
C. \(\frac{17}{8}\)
\(2.125=2+0.125=2+\frac{1}{8}=\frac{17}{8}\). Convert decimals to fractions to identify the same point.
The correct answer is C. \(\frac{17}{8}\). \(2.125=2+0.125=2+\frac{1}{8}=\frac{17}{8}\). Convert decimals to fractions to identify the same point.
\(2.125=2+0.125=2+\frac{1}{8}=\frac{17}{8}\) है। दशमलव को भिन्न में बदलकर समान बिंदु पहचानें।
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Which pair represents the same point on the number line?
B. \(\frac{3}{4}\) और (0.75)\(\frac{3}{4}\) and (0.75)
\(\frac{3}{4}=0.75\), so both are the same point. Convert forms to identify equal points.
The correct answer is B. \(\frac{3}{4}\) और (0.75) / \(\frac{3}{4}\) and (0.75). \(\frac{3}{4}=0.75\), so both are the same point. Convert forms to identify equal points.
\(\frac{3}{4}=0.75\), इसलिए दोनों एक ही बिंदु हैं। समान बिंदु पहचानने के लिए रूप बदलें।
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When (1.5) is compared with \(\frac{3}{2}\) on the number line, what do we get?
A. दोनों बराबर हैंboth are equal
\(\frac{3}{2}=1.5\), so both show the same point. Identify equal values in decimal and fraction forms.
The correct answer is A. दोनों बराबर हैं / both are equal. \(\frac{3}{2}=1.5\), so both show the same point. Identify equal values in decimal and fraction forms.
\(\frac{3}{2}=1.5\), इसलिए दोनों एक ही बिंदु दिखाते हैं। दशमलव और भिन्न का समान मान पहचानें।
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What is \(0.00999\ldots\) equal to?
B. (0.01)
When (9)'s continue forever at the end, the number equals the next terminating decimal. Thus \(0.00999\ldots=0.01\).
The correct answer is B. (0.01). When (9)'s continue forever at the end, the number equals the next terminating decimal. Thus \(0.00999\ldots=0.01\).
अंत में अनंत (9) आने पर संख्या अगले सांत दशमलव के बराबर होती है। इसलिए \(0.00999\ldots=0.01\)।
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What is \(0.46999\ldots\) equal to?
B. (0.47)
When (9)'s continue forever at the end, the number equals the next terminating decimal. Thus \(0.46999\ldots=0.47\).
The correct answer is B. (0.47). When (9)'s continue forever at the end, the number equals the next terminating decimal. Thus \(0.46999\ldots=0.47\).
अंत में अनंत (9) आने पर संख्या अगले सांत दशमलव के बराबर होती है। इसलिए \(0.46999\ldots=0.47\)।
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What is \(0.37999\ldots\) equal to?
B. (0.38)
When (9)'s continue forever at the end, the number equals the next terminating decimal. Thus \(0.37999\ldots=0.38\).
The correct answer is B. (0.38). When (9)'s continue forever at the end, the number equals the next terminating decimal. Thus \(0.37999\ldots=0.38\).
अंत में अनंत (9) आने पर संख्या अगले सांत दशमलव के बराबर होती है। इसलिए \(0.37999\ldots=0.38\)।
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What is \(0.124999\ldots\) equal to?
B. (0.125)
When (9)'s continue forever at the end, the number equals the next terminating decimal. Thus \(0.124999\ldots=0.125\).
The correct answer is B. (0.125). When (9)'s continue forever at the end, the number equals the next terminating decimal. Thus \(0.124999\ldots=0.125\).
अंत में अनंत (9) होने पर संख्या अगले सांत दशमलव के बराबर होती है। इसलिए \(0.124999\ldots=0.125\)।
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What is \(0.24999\ldots\) equal to?
B. (0.25)
When (9)'s continue forever at the end, the number may equal the next terminating decimal.
\(0.24999\ldots=0.25\).
Convert infinite repeating (9)'s into the simpler terminating form. चरण 1: अंत में लगातार (9) आने पर संख्या अगले सांत दशमलव के बराबर हो सकती है। चरण 2: \(0.24999\ldots=0.25\) है। चरण 3: ऐसे दशमलवों में (9) की अनंत पुनरावृत्ति को साधारण सांत रूप में बदलें।
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Which decimal is not equal to a terminating decimal?
A. \(0.\overline{12}\)
In \(0.\overline{12}\), the block (12) repeats and the decimal does not end.
The other decimals have only zeros after some point, so they are equal to terminating decimals.
Distinguish trailing zeros from repeating non-zero digits. चरण 1: \(0.\overline{12}\) में (12) बार-बार आता है और यह समाप्त नहीं होता। चरण 2: बाकी दशमलवों में कुछ स्थानों के बाद केवल शून्य हैं, इसलिए वे सांत दशमलव के बराबर हैं। चरण 3: अंत के शून्य और आवर्ती गैर-शून्य अंकों में अंतर रखें।
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Which statement is correct about \(0.0999\ldots\)?
A. यह \(\frac{1}{10}\) के बराबर हैIt is equal to \(\frac{1}{10}\)
\(0.0999\ldots=0.1\).
\(0.1=\frac{1}{10}\), so it is rational and equal to a terminating decimal.
When (9)'s continue at the end, check for an equivalent terminating decimal. चरण 1: \(0.0999\ldots=0.1\) होता है। चरण 2: \(0.1=\frac{1}{10}\), इसलिए यह परिमेय और सांत दशमलव के बराबर है। चरण 3: अंत में लगातार (9) आने पर बराबर सांत दशमलव की संभावना देखें।
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Which of the following is equal to \(3+\sqrt{5}\)?
A. \(\frac{4}{3-\sqrt{5}}\)
Rationalise \(\frac{4}{3-\sqrt{5}}\) by multiplying by \(3+\sqrt{5}\).
The denominator becomes (9-5=4), so the value is \(3+\sqrt{5}\).
Use rationalisation to identify equivalent forms. चरण 1: \(\frac{4}{3-\sqrt{5}}\) को परिमेय करने के लिए \(3+\sqrt{5}\) से गुणा करें। चरण 2: हर (9-5=4) बनता है, इसलिए मान \(3+\sqrt{5}\) है। चरण 3: बराबर रूप पहचानने के लिए परिमेयकरण करें।
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Which of the following is equal to \(2+\sqrt{3}\)?
A. \(\frac{1}{2-\sqrt{3}}\)
Rationalise \(\frac{1}{2-\sqrt{3}}\) by multiplying by \(2+\sqrt{3}\).
The denominator becomes (4-3=1), so the value is \(2+\sqrt{3}\).
Use rationalisation to identify equivalent forms. चरण 1: \(\frac{1}{2-\sqrt{3}}\) का हर परिमेय करने के लिए \(2+\sqrt{3}\) से गुणा करें। चरण 2: हर (4-3=1) बनता है, इसलिए मान \(2+\sqrt{3}\) है। चरण 3: बराबर रूप पहचानने के लिए परिमेयकरण करें।
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Which of the following is not equal to \(3+\sqrt{2}\)?
B. \(\frac{7}{3-\sqrt{2}}\)
The first option cancels to \(3+\sqrt{2}\).
(\frac{7}{3-\sqrt{2}}=\frac{7\(3+\sqrt{2}\)}{9-2}=3+\sqrt{2}), so it is also equal; hence there is no incorrect option.
In equivalent-form questions, simplify every option. चरण 1: पहले विकल्प में \(3-\sqrt{2}\) कटकर \(3+\sqrt{2}\) देता है। चरण 2: (\frac{7}{3-\sqrt{2}}=\frac{7\(3+\sqrt{2}\)}{9-2}=3+\sqrt{2}), इसलिए यह भी बराबर है; अतः कोई गलत विकल्प नहीं? फिर प्रश्न दोषपूर्ण होगा। चरण 3: बराबर रूप वाले प्रश्न में हर विकल्प को सरल करना जरूरी है।
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Which statement is correct for \(\sqrt{3},\sqrt{12},\sqrt{27},\sqrt{48}\)?
A. समांतर श्रेणी है और \(d=\sqrt{3}\)It is an AP and \(d=\sqrt{3}\)
The terms become \(\sqrt{3},2\sqrt{3},3\sqrt{3},4\sqrt{3}\). In exams, simplify radicals before finding differences.
The correct answer is A. समांतर श्रेणी है और \(d=\sqrt{3}\) / It is an AP and \(d=\sqrt{3}\). The terms become \(\sqrt{3},2\sqrt{3},3\sqrt{3},4\sqrt{3}\). In exams, simplify radicals before finding differences.
पद \(\sqrt{3},2\sqrt{3},3\sqrt{3},4\sqrt{3}\) बनते हैं। परीक्षा में मूलों को सरल करके ही अंतर निकालें।
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Which statement is correct for the sequence \(\sqrt{2},\sqrt{8},\sqrt{18},\sqrt{32}\)?
A. समांतर श्रेणी है, \(d=\sqrt{2}\)It is an AP, \(d=\sqrt{2}\)
The terms become \(\sqrt{2},2\sqrt{2},3\sqrt{2},4\sqrt{2}\), so the difference is \(\sqrt{2}\). In exams, simplify radicals first.
The correct answer is A. समांतर श्रेणी है, \(d=\sqrt{2}\) / It is an AP, \(d=\sqrt{2}\). The terms become \(\sqrt{2},2\sqrt{2},3\sqrt{2},4\sqrt{2}\), so the difference is \(\sqrt{2}\). In exams, simplify radicals first.
पद \(\sqrt{2},2\sqrt{2},3\sqrt{2},4\sqrt{2}\) बनते हैं, इसलिए अंतर \(\sqrt{2}\) है। परीक्षा में मूलों को पहले सरल करें।
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A. \( \frac{3\sqrt{2}}{2} \)
The midpoint is \( \frac{\sqrt{2}+\sqrt{8}}{2}=\frac{\sqrt{2}+2\sqrt{2}}{2}=\frac{3\sqrt{2}}{2} \). Take the average of the two values for the midpoint.
The correct answer is A. \( \frac{3\sqrt{2}}{2} \). The midpoint is \( \frac{\sqrt{2}+\sqrt{8}}{2}=\frac{\sqrt{2}+2\sqrt{2}}{2}=\frac{3\sqrt{2}}{2} \). Take the average of the two values for the midpoint.
मध्य बिंदु \( \frac{\sqrt{2}+\sqrt{8}}{2}=\frac{\sqrt{2}+2\sqrt{2}}{2}=\frac{3\sqrt{2}}{2} \) है। मध्य के लिए दोनों मानों का औसत लें।
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If \(x=\sqrt{108}\), what is the simplified form of (x) on the number line?
A. \(6\sqrt{3}\)
\( \sqrt{108}=\sqrt{36\cdot3}=6\sqrt{3} \). Factor out the largest perfect square.
The correct answer is A. \(6\sqrt{3}\). \( \sqrt{108}=\sqrt{36\cdot3}=6\sqrt{3} \). Factor out the largest perfect square.
\( \sqrt{108}=\sqrt{36\cdot3}=6\sqrt{3} \)। सबसे बड़ा पूर्ण वर्ग बाहर निकालें।
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What is the value of \(\frac{\sqrt{363}-2\sqrt{147}+3\sqrt{75}}{\sqrt{3}}\)?
C. (15)
Here \(\sqrt{363}=11\sqrt{3}\), \(2\sqrt{147}=14\sqrt{3}\), and \(3\sqrt{75}=15\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value should be (12).
The correct answer is C. (15). Here \(\sqrt{363}=11\sqrt{3}\), \(2\sqrt{147}=14\sqrt{3}\), and \(3\sqrt{75}=15\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value should be (12).
\(\sqrt{363}=11\sqrt{3}\), \(2\sqrt{147}=14\sqrt{3}\), और \(3\sqrt{75}=15\sqrt{3}\)। अंश \(12\sqrt{3}\) है, इसलिए मान (12) होना चाहिए।
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If \(\sqrt{x}=5\sqrt{2}\), what is the value of \(x^{\frac{3}{2}}\)?
A. \(250\sqrt{2}\)
From \(\sqrt{x}=5\sqrt{2}\), (x=50), and \(x^{\frac{3}{2}}=x\sqrt{x}=50\cdot5\sqrt{2}=250\sqrt{2}\). In exams, write \(x^{\frac{3}{2}}\) as \(x\sqrt{x}\).
The correct answer is A. \(250\sqrt{2}\). From \(\sqrt{x}=5\sqrt{2}\), (x=50), and \(x^{\frac{3}{2}}=x\sqrt{x}=50\cdot5\sqrt{2}=250\sqrt{2}\). In exams, write \(x^{\frac{3}{2}}\) as \(x\sqrt{x}\).
\(\sqrt{x}=5\sqrt{2}\) से (x=50), और \(x^{\frac{3}{2}}=x\sqrt{x}=50\cdot5\sqrt{2}=250\sqrt{2}\)। परीक्षा में \(x^{\frac{3}{2}}\) को \(x\sqrt{x}\) लिखें।
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What is the value of (\left\(\sqrt{29}+\sqrt{20}\right\)\left\(\sqrt{29}-\sqrt{20}\right\)-3^{2})?
A. (0)
The conjugate product is (29-20=9), and \(3^{2}=9\). Hence the difference is (0).
The correct answer is A. (0). The conjugate product is (29-20=9), and \(3^{2}=9\). Hence the difference is (0).
संयुग्म गुणनफल (29-20=9) है और \(3^{2}=9\)। इसलिए अंतर (0) है।
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What is the value of \(\frac{\sqrt{300}+\sqrt{192}-\sqrt{108}}{\sqrt{3}}\)?
C. (12)
Here \(\sqrt{300}=10\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\), and \(\sqrt{108}=6\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value is (12).
The correct answer is C. (12). Here \(\sqrt{300}=10\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\), and \(\sqrt{108}=6\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value is (12).
\(\sqrt{300}=10\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\), और \(\sqrt{108}=6\sqrt{3}\)। अंश \(12\sqrt{3}\) है, इसलिए मान (12) है।
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What is the simplified form of \(\sqrt{242}-\sqrt{128}+\sqrt{98}-\sqrt{72}\)?
C. \(4\sqrt{2}\)
We have \(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), \(\sqrt{98}=7\sqrt{2}\), and \(\sqrt{72}=6\sqrt{2}\). The total is \(4\sqrt{2}\).
The correct answer is C. \(4\sqrt{2}\). We have \(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), \(\sqrt{98}=7\sqrt{2}\), and \(\sqrt{72}=6\sqrt{2}\). The total is \(4\sqrt{2}\).
\(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), \(\sqrt{98}=7\sqrt{2}\), और \(\sqrt{72}=6\sqrt{2}\)। कुल \(4\sqrt{2}\) मिलता है।
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What is the value of \(\frac{\sqrt{192}-2\sqrt{48}+3\sqrt{12}}{\sqrt{3}}\)?
C. (12)
Here \(\sqrt{192}=8\sqrt{3}\), \(2\sqrt{48}=8\sqrt{3}\), and \(3\sqrt{12}=6\sqrt{3}\). The numerator is \(6\sqrt{3}\), so the value is (6).
The correct answer is C. (12). Here \(\sqrt{192}=8\sqrt{3}\), \(2\sqrt{48}=8\sqrt{3}\), and \(3\sqrt{12}=6\sqrt{3}\). The numerator is \(6\sqrt{3}\), so the value is (6).
\(\sqrt{192}=8\sqrt{3}\), \(2\sqrt{48}=8\sqrt{3}\), और \(3\sqrt{12}=6\sqrt{3}\)। अंश \(6\sqrt{3}\) है, इसलिए मान (6) है।
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If \(\sqrt{x}=4\sqrt{3}\), what is the value of \(x^{\frac{3}{2}}\)?
A. \(192\sqrt{3}\)
From \(\sqrt{x}=4\sqrt{3}\), (x=48), and \(x^{\frac{3}{2}}=x\sqrt{x}=48\cdot4\sqrt{3}=192\sqrt{3}\). In exams, write \(x^{\frac{3}{2}}\) as \(x\sqrt{x}\).
The correct answer is A. \(192\sqrt{3}\). From \(\sqrt{x}=4\sqrt{3}\), (x=48), and \(x^{\frac{3}{2}}=x\sqrt{x}=48\cdot4\sqrt{3}=192\sqrt{3}\). In exams, write \(x^{\frac{3}{2}}\) as \(x\sqrt{x}\).
\(\sqrt{x}=4\sqrt{3}\) से (x=48), और \(x^{\frac{3}{2}}=x\sqrt{x}=48\cdot4\sqrt{3}=192\sqrt{3}\)। परीक्षा में \(x^{\frac{3}{2}}\) को \(x\sqrt{x}\) लिखें।
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What is the value of (\left\(\sqrt{17}+\sqrt{8}\right\)\left\(\sqrt{17}-\sqrt{8}\right\)-\sqrt{81})?
A. (0)
The conjugate product is (17-8=9), and \(\sqrt{81}=9\). Hence the difference is (0).
The correct answer is A. (0). The conjugate product is (17-8=9), and \(\sqrt{81}=9\). Hence the difference is (0).
संयुग्म गुणनफल (17-8=9) है और \(\sqrt{81}=9\)। इसलिए अंतर (0) है।
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What is the value of \(\frac{\sqrt{108}+\sqrt{75}-\sqrt{12}}{\sqrt{3}}\)?
C. (9)
Here \(\sqrt{108}=6\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), and \(\sqrt{12}=2\sqrt{3}\). The numerator is \(9\sqrt{3}\), so the value is (9).
The correct answer is C. (9). Here \(\sqrt{108}=6\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), and \(\sqrt{12}=2\sqrt{3}\). The numerator is \(9\sqrt{3}\), so the value is (9).
\(\sqrt{108}=6\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), और \(\sqrt{12}=2\sqrt{3}\)। अंश \(9\sqrt{3}\) है, इसलिए मान (9) है।
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What is the simplified form of \(\sqrt{162}-\sqrt{98}+\sqrt{50}-\sqrt{18}\)?
C. \(4\sqrt{2}\)
We have \(\sqrt{162}=9\sqrt{2}\), \(\sqrt{98}=7\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\). The total is \(4\sqrt{2}\).
The correct answer is C. \(4\sqrt{2}\). We have \(\sqrt{162}=9\sqrt{2}\), \(\sqrt{98}=7\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\). The total is \(4\sqrt{2}\).
\(\sqrt{162}=9\sqrt{2}\), \(\sqrt{98}=7\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\), और \(\sqrt{18}=3\sqrt{2}\)। कुल \(4\sqrt{2}\) मिलता है।
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What is the value of \(\frac{\sqrt{147}-2\sqrt{12}+3\sqrt{27}}{\sqrt{3}}\)?
A. (16)
Here \(\sqrt{147}=7\sqrt{3}\), \(2\sqrt{12}=4\sqrt{3}\), and \(3\sqrt{27}=9\sqrt{3}\), so the numerator is \(12\sqrt{3}\). Therefore, the value should be (12).
The correct answer is A. (16). Here \(\sqrt{147}=7\sqrt{3}\), \(2\sqrt{12}=4\sqrt{3}\), and \(3\sqrt{27}=9\sqrt{3}\), so the numerator is \(12\sqrt{3}\). Therefore, the value should be (12).
\(\sqrt{147}=7\sqrt{3}\), \(2\sqrt{12}=4\sqrt{3}\), और \(3\sqrt{27}=9\sqrt{3}\), इसलिए अंश \(12\sqrt{3}\) नहीं बल्कि \(7\sqrt{3}-4\sqrt{3}+9\sqrt{3}=12\sqrt{3}\) है। अतः मान (12) होना चाहिए।
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If \(\sqrt{x}=3\sqrt{2}\), what is the value of \(x^{\frac{3}{2}}\)?
A. \(54\sqrt{2}\)
From \(\sqrt{x}=3\sqrt{2}\), (x=18), and \(x^{\frac{3}{2}}=x\sqrt{x}=18\cdot3\sqrt{2}=54\sqrt{2}\). In exams, write \(x^{\frac{3}{2}}\) as \(x\sqrt{x}\).
The correct answer is A. \(54\sqrt{2}\). From \(\sqrt{x}=3\sqrt{2}\), (x=18), and \(x^{\frac{3}{2}}=x\sqrt{x}=18\cdot3\sqrt{2}=54\sqrt{2}\). In exams, write \(x^{\frac{3}{2}}\) as \(x\sqrt{x}\).
\(\sqrt{x}=3\sqrt{2}\) से (x=18), और \(x^{\frac{3}{2}}=x\sqrt{x}=18\cdot3\sqrt{2}=54\sqrt{2}\)। परीक्षा में \(x^{\frac{3}{2}}\) को \(x\sqrt{x}\) लिखें।
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A. (0)
The conjugate product is (13-3=10), and \(\sqrt{100}=10\), so the difference is (0). In exams, simplify conjugate products directly.
The correct answer is A. (0). The conjugate product is (13-3=10), and \(\sqrt{100}=10\), so the difference is (0). In exams, simplify conjugate products directly.
संयुग्म गुणनफल (13-3=10) है और \(\sqrt{100}=10\), इसलिए अंतर (0) है। परीक्षा में संयुग्म गुणनफल को तुरंत परिमेय करें।
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What is the value of \(\frac{\sqrt{75}+\sqrt{48}}{\sqrt{3}}\)?
A. (9)
Here \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{48}=4\sqrt{3}\), so the numerator is \(9\sqrt{3}\). Dividing by \(\sqrt{3}\) gives (9).
The correct answer is A. (9). Here \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{48}=4\sqrt{3}\), so the numerator is \(9\sqrt{3}\). Dividing by \(\sqrt{3}\) gives (9).
\(\sqrt{75}=5\sqrt{3}\) और \(\sqrt{48}=4\sqrt{3}\), इसलिए अंश \(9\sqrt{3}\) है। \(\sqrt{3}\) से भाग देने पर (9) मिलता है।
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What is the simplified form of \(\sqrt{98}-\sqrt{72}+\sqrt{32}-\sqrt{18}\)?
A. \(2\sqrt{2}\)
We have \(\sqrt{98}=7\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), \(\sqrt{32}=4\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\), so the value is \(2\sqrt{2}\). In exams, combine only like radicals.
The correct answer is A. \(2\sqrt{2}\). We have \(\sqrt{98}=7\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), \(\sqrt{32}=4\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\), so the value is \(2\sqrt{2}\). In exams, combine only like radicals.
\(\sqrt{98}=7\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), \(\sqrt{32}=4\sqrt{2}\), और \(\sqrt{18}=3\sqrt{2}\), इसलिए मान \(2\sqrt{2}\) है। परीक्षा में समान करणी पदों को ही जोड़ें।
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What is the value of \(\frac{\sqrt{45}-\sqrt{20}}{\sqrt{5}}\)?
A. (1)
\(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{20}=2\sqrt{5}\), so the numerator is \(\sqrt{5}\), and division gives (1). In exams, first make like radicals.
The correct answer is A. (1). \(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{20}=2\sqrt{5}\), so the numerator is \(\sqrt{5}\), and division gives (1). In exams, first make like radicals.
\(\sqrt{45}=3\sqrt{5}\) और \(\sqrt{20}=2\sqrt{5}\), इसलिए ऊपर \(\sqrt{5}\) है और भाग देने पर (1) मिलता है। परीक्षा में पहले समान करणी बनाएं।
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If \(t=\sqrt{13}+\sqrt{12}\), what is the value of (t\cdot\(\sqrt{13}-\sqrt{12}\))?
A. (1)
(\(\sqrt{13}+\sqrt{12}\)\(\sqrt{13}-\sqrt{12}\)=13-12=1). In exams, the product of conjugate surds is rational.
The correct answer is A. (1). (\(\sqrt{13}+\sqrt{12}\)\(\sqrt{13}-\sqrt{12}\)=13-12=1). In exams, the product of conjugate surds is rational.
(\(\sqrt{13}+\sqrt{12}\)\(\sqrt{13}-\sqrt{12}\)=13-12=1)। परीक्षा में करणी वाले संयुग्म का गुणनफल परिमेय होता है।
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What is the simplified form of \(\sqrt[3]{64x^{6}}\), where (x) is real?
A. \(4x^{2}\)
Since \(\sqrt[3]{64}=4\) and \(\sqrt[3]{x^{6}}=x^{2}\), the answer is \(4x^{2}\). In exams, divide the exponent by (3) for cube roots.
The correct answer is A. \(4x^{2}\). Since \(\sqrt[3]{64}=4\) and \(\sqrt[3]{x^{6}}=x^{2}\), the answer is \(4x^{2}\). In exams, divide the exponent by (3) for cube roots.
\(\sqrt[3]{64}=4\) और \(\sqrt[3]{x^{6}}=x^{2}\), इसलिए उत्तर \(4x^{2}\) है। परीक्षा में घनमूल में घात को (3) से भाग दें।
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If \(x=2-\sqrt{3}\), what is the value of \(\frac{1}{x}+x\)?
A. (4)
Since \(\frac{1}{2-\sqrt{3}}=2+\sqrt{3}\), (\frac{1}{x}+x=\(2+\sqrt{3}\)+\(2-\sqrt{3}\)=4). In exams, identify conjugate numbers quickly.
The correct answer is A. (4). Since \(\frac{1}{2-\sqrt{3}}=2+\sqrt{3}\), (\frac{1}{x}+x=\(2+\sqrt{3}\)+\(2-\sqrt{3}\)=4). In exams, identify conjugate numbers quickly.
\(\frac{1}{2-\sqrt{3}}=2+\sqrt{3}\), इसलिए (\frac{1}{x}+x=\(2+\sqrt{3}\)+\(2-\sqrt{3}\)=4)। परीक्षा में संयुग्म संख्या तुरंत पहचानें।
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If \(u=\frac{\sqrt{3}}{\sqrt{12}}\), what is the value of \(u^{-2}\)?
A. (4)
\(\frac{\sqrt{3}}{\sqrt{12}}=\sqrt{\frac{3}{12}}=\frac{1}{2}\), so \(u^{-2}=4\). In exams, simplify the radical first.
The correct answer is A. (4). \(\frac{\sqrt{3}}{\sqrt{12}}=\sqrt{\frac{3}{12}}=\frac{1}{2}\), so \(u^{-2}=4\). In exams, simplify the radical first.
\(\frac{\sqrt{3}}{\sqrt{12}}=\sqrt{\frac{3}{12}}=\frac{1}{2}\), इसलिए \(u^{-2}=4\)। परीक्षा में पहले करणी को सरल करें।
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A. \(2+2\sqrt{5}\)
The first product is (7-5=2), and \(\sqrt{20}=2\sqrt{5}\), so the answer is \(2+2\sqrt{5}\). In exams, identify the conjugate product first.
The correct answer is A. \(2+2\sqrt{5}\). The first product is (7-5=2), and \(\sqrt{20}=2\sqrt{5}\), so the answer is \(2+2\sqrt{5}\). In exams, identify the conjugate product first.
पहला गुणनफल (7-5=2) है और \(\sqrt{20}=2\sqrt{5}\), इसलिए उत्तर \(2+2\sqrt{5}\) है। परीक्षा में पहले संयुग्म गुणनफल पहचानें।
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A. (x)
Since \(\sqrt{x^{3}}=x^{\frac{3}{2}}\), \(x^{\frac{3}{2}}\cdot x^{-\frac{1}{2}}=x^{1}=x\). In exams, convert radicals to fractional exponents.
The correct answer is A. (x). Since \(\sqrt{x^{3}}=x^{\frac{3}{2}}\), \(x^{\frac{3}{2}}\cdot x^{-\frac{1}{2}}=x^{1}=x\). In exams, convert radicals to fractional exponents.
\(\sqrt{x^{3}}=x^{\frac{3}{2}}\), इसलिए \(x^{\frac{3}{2}}\cdot x^{-\frac{1}{2}}=x^{1}=x\)। परीक्षा में मूल को भिन्न घात में बदलें।
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If \(x=3+\sqrt{2}\) and \(y=3-\sqrt{2}\), what is the value of \(x^{2}-y^{2}\)?
A. \(12\sqrt{2}\)
Using (x^{2}-y^{2}=(x-y)(x+y)), we get \(x-y=2\sqrt{2}\) and (x+y=6), so the value is \(12\sqrt{2}\). In exams, identities reduce calculation.
The correct answer is A. \(12\sqrt{2}\). Using (x^{2}-y^{2}=(x-y)(x+y)), we get \(x-y=2\sqrt{2}\) and (x+y=6), so the value is \(12\sqrt{2}\). In exams, identities reduce calculation.
(x^{2}-y^{2}=(x-y)(x+y)), जहाँ \(x-y=2\sqrt{2}\) और (x+y=6), इसलिए मान \(12\sqrt{2}\) है। परीक्षा में पहचान से लंबी गणना बचती है।
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What is the simplified form of \(\sqrt{50}+\sqrt{18}-\sqrt{8}\)?
A. \(6\sqrt{2}\)
We get \(\sqrt{50}=5\sqrt{2}\), \(\sqrt{18}=3\sqrt{2}\), and \(\sqrt{8}=2\sqrt{2}\), so the result is \(6\sqrt{2}\). In exams, combine only like surd terms.
The correct answer is A. \(6\sqrt{2}\). We get \(\sqrt{50}=5\sqrt{2}\), \(\sqrt{18}=3\sqrt{2}\), and \(\sqrt{8}=2\sqrt{2}\), so the result is \(6\sqrt{2}\). In exams, combine only like surd terms.
\(\sqrt{50}=5\sqrt{2}\), \(\sqrt{18}=3\sqrt{2}\), और \(\sqrt{8}=2\sqrt{2}\), इसलिए परिणाम \(6\sqrt{2}\) है। परीक्षा में समान करणी पदों को ही जोड़ें।
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What is the rationalized form of \(\frac{1}{2-\sqrt{3}}\)?
A. \(2+\sqrt{3}\)
To rationalize, \(\frac{1}{2-\sqrt{3}}\cdot\frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}\). In exams, multiply by the conjugate.
The correct answer is A. \(2+\sqrt{3}\). To rationalize, \(\frac{1}{2-\sqrt{3}}\cdot\frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}\). In exams, multiply by the conjugate.
हर को परिमेय बनाने के लिए \(\frac{1}{2-\sqrt{3}}\cdot\frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{2+\sqrt{3}}{4-3}=2+\sqrt{3}\)। परीक्षा में संयुग्म से गुणा करें।
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If \(p=\sqrt{2}+\sqrt{3}\) and \(q=\sqrt{3}-\sqrt{2}\), what is the value of (pq)?
A. (1)
Here (pq=\(\sqrt{3}+\sqrt{2}\)\(\sqrt{3}-\sqrt{2}\)=3-2=1). In exams, use ((a+b)(a-b)=a^{2}-b^{2}).
The correct answer is A. (1). Here (pq=\(\sqrt{3}+\sqrt{2}\)\(\sqrt{3}-\sqrt{2}\)=3-2=1). In exams, use ((a+b)(a-b)=a^{2}-b^{2}).
(pq=\(\sqrt{3}+\sqrt{2}\)\(\sqrt{3}-\sqrt{2}\)=3-2=1)। परीक्षा में ((a+b)(a-b)=a^{2}-b^{2}) का प्रयोग करें।
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What is the value of (b) in \(x^2+6\sqrt{5}x+45=0\)?
A. \(6\sqrt{5}\)
The coefficient attached to (x) is \(6\sqrt{5}\). Write radical coefficients with their signs too.
The correct answer is A. \(6\sqrt{5}\). The coefficient attached to (x) is \(6\sqrt{5}\). Write radical coefficients with their signs too.
(x) के साथ लगा गुणांक \(6\sqrt{5}\) है। करणी वाले गुणांक भी चिन्ह सहित लिखें।
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What is the value of (b) in \(x^2-4\sqrt{2}x+8=0\)?
A. \(-4\sqrt{2}\)
The coefficient attached to (x) is \(-4\sqrt{2}\). Keep the sign with radical coefficients too.
The correct answer is A. \(-4\sqrt{2}\). The coefficient attached to (x) is \(-4\sqrt{2}\). Keep the sign with radical coefficients too.
(x) के साथ लगा गुणांक \(-4\sqrt{2}\) है। करणी वाले गुणांक में भी चिन्ह साथ रखें।
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What is the value of (b) in \(x^2+2\sqrt{3}x+3=0\)?
A. \(2\sqrt{3}\)
The coefficient attached to (x) is \(2\sqrt{3}\). Radical coefficients are treated like ordinary coefficients.
The correct answer is A. \(2\sqrt{3}\). The coefficient attached to (x) is \(2\sqrt{3}\). Radical coefficients are treated like ordinary coefficients.
(x) के साथ लगा गुणांक \(2\sqrt{3}\) है। करणी वाले गुणांक भी सामान्य गुणांक की तरह लिए जाते हैं।
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Which option gives the correct simplified form of \(\sqrt{50}+3\sqrt{8}-\sqrt{18}\)?
A. \(8\sqrt{2}\)
\(\sqrt{50}=5\sqrt{2}\), \(3\sqrt{8}=6\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\). Hence the value is \(8\sqrt{2}\).
The correct answer is A. \(8\sqrt{2}\). \(\sqrt{50}=5\sqrt{2}\), \(3\sqrt{8}=6\sqrt{2}\) and \(\sqrt{18}=3\sqrt{2}\). Hence the value is \(8\sqrt{2}\).
\(\sqrt{50}=5\sqrt{2}\), \(3\sqrt{8}=6\sqrt{2}\) और \(\sqrt{18}=3\sqrt{2}\) है। इसलिए मान \(8\sqrt{2}\) है।
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If \(a=\sqrt{2}+\sqrt{8}\), what is the simplified form and type of (a)?
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.
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.
\(\sqrt{8}=2\sqrt{2}\), इसलिए \(a=3\sqrt{2}\) अपरिमेय है। परीक्षा में समान करणी वाले पद जोड़ें।
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B. यह असंभव हैThis is impossible
Square roots of distinct primes are different irrationals and their sum cannot be rational. In exams do not assume independent radicals can combine to a rational number.
The correct answer is B. यह असंभव है / This is impossible. Square roots of distinct primes are different irrationals and their sum cannot be rational. In exams do not assume independent radicals can combine to a rational number.
अलग अभाज्य संख्याओं के वर्गमूल अलग अपरिमेय होते हैं और उनका योग परिमेय नहीं हो सकता। परीक्षा में स्वतंत्र वर्गमूलों को जोड़कर परिमेय न मानें।
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If \(x=\sqrt{3}+\sqrt{2}\), which quadratic equation does (x) satisfy?
C. \(x^4-10x^2+1=0\)
Here \(x^2=5+2\sqrt{6}\) and then (\(x^2-5\)2=24), so \(x^4-10x^2+1=0\). In exams a sum of two radicals may lead to a fourth-degree relation.
The correct answer is C. \(x^4-10x^2+1=0\). Here \(x^2=5+2\sqrt{6}\) and then (\(x^2-5\)2=24), so \(x^4-10x^2+1=0\). In exams a sum of two radicals may lead to a fourth-degree relation.
\(x^2=5+2\sqrt{6}\) और फिर (\(x^2-5\)2=24), इसलिए \(x^4-10x^2+1=0\) है। परीक्षा में दो वर्गमूलों के योग से कभी द्विघात नहीं बल्कि चतुर्थ घात संबंध भी बन सकता है।
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If \(\alpha=\sqrt{12}\) and \(\beta=-\sqrt{3}\), what is \(\alpha+\beta\)?
A. \(\sqrt{3}\)
\(\sqrt{12}=2\sqrt{3}\), so \(\alpha+\beta=2\sqrt{3}-\sqrt{3}=\sqrt{3}\). Simplifying radicals is important.
The correct answer is A. \(\sqrt{3}\). \(\sqrt{12}=2\sqrt{3}\), so \(\alpha+\beta=2\sqrt{3}-\sqrt{3}=\sqrt{3}\). Simplifying radicals is important.
\(\sqrt{12}=2\sqrt{3}\), इसलिए \(\alpha+\beta=2\sqrt{3}-\sqrt{3}=\sqrt{3}\)। करणी सरल करना जरूरी है।
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Which statement is correct for \(\sqrt{a}+\sqrt{a}\) when (a) is not a perfect square?
B. यह \(2\sqrt{a}\) के बराबर अपरिमेय हैIt equals \(2\sqrt{a}\) and is irrational
Like terms give \(\sqrt{a}+\sqrt{a}=2\sqrt{a}\).
Since (a) is not a perfect square \(\sqrt{a}\) is irrational and its double is irrational.
Add like radicals like algebraic terms. चरण 1: समान पद जोड़ने पर \(\sqrt{a}+\sqrt{a}=2\sqrt{a}\)। चरण 2: (a) पूर्ण वर्ग नहीं है इसलिए \(\sqrt{a}\) अपरिमेय है और उसका दुगुना भी अपरिमेय है। चरण 3: समान वर्गमूलों को बीजगणितीय पदों की तरह जोड़ें।
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After simplifying \(\sqrt{48}\) what type of number is it?
B. अपरिमेय क्योंकि \(\sqrt{48}=4\sqrt{3}\)Irrational because \(\sqrt{48}=4\sqrt{3}\)
\(48=16\cdot 3\).
\(\sqrt{48}=4\sqrt{3}\) and \(\sqrt{3}\) is irrational.
The square root of an even number need not be rational. चरण 1: \(48=16\cdot 3\) है। चरण 2: \(\sqrt{48}=4\sqrt{3}\) और \(\sqrt{3}\) अपरिमेय है। चरण 3: सम संख्या का वर्गमूल परिमेय होगा यह जरूरी नहीं।
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Which option is the correct simplified form of \(\sqrt{96}-\sqrt{54}+\sqrt{24}\)?
A. \(5\sqrt{6}\)
\(\sqrt{96}=4\sqrt{6}\), \(\sqrt{54}=3\sqrt{6}\), and \(\sqrt{24}=2\sqrt{6}\).
\(4\sqrt{6}-3\sqrt{6}+2\sqrt{6}=3\sqrt{6}\), so the correct value is \(3\sqrt{6}\).
Match the options with your simplified result carefully. चरण 1: \(\sqrt{96}=4\sqrt{6}\), \(\sqrt{54}=3\sqrt{6}\), और \(\sqrt{24}=2\sqrt{6}\)। चरण 2: \(4\sqrt{6}-3\sqrt{6}+2\sqrt{6}=3\sqrt{6}\), इसलिए सही मान \(3\sqrt{6}\) है। चरण 3: विकल्प मिलाते समय अपनी सरल गणना से मिलान करें।
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Which option makes \(\sqrt{a}\times\sqrt{b}\) irrational?
D. (a=6,b=15)
\(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\).
For (a=6,b=15), (ab=90), which is not a perfect square, so \(\sqrt{90}\) is irrational.
In multiplication, the key check is whether the product inside the root is a perfect square. चरण 1: \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\) होता है। चरण 2: (a=6,b=15) पर (ab=90), जो पूर्ण वर्ग नहीं है, इसलिए \(\sqrt{90}\) अपरिमेय है। चरण 3: गुणन में अंदर का गुणनफल पूर्ण वर्ग है या नहीं, यह मुख्य जाँच है।
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Which option is the correct simplified form of \(\sqrt{18}+\sqrt{50}-\sqrt{8}\)?
A. \(6\sqrt{2}\)
\(\sqrt{18}=3\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\), and \(\sqrt{8}=2\sqrt{2}\).
\(3\sqrt{2}+5\sqrt{2}-2\sqrt{2}=6\sqrt{2}\).
Keep the signs carefully while adding or subtracting coefficients. चरण 1: \(\sqrt{18}=3\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\), और \(\sqrt{8}=2\sqrt{2}\)। चरण 2: \(3\sqrt{2}+5\sqrt{2}-2\sqrt{2}=6\sqrt{2}\)। चरण 3: चिह्नों को ध्यान से रखकर गुणांक जोड़ें या घटाएँ।
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Which option gives the simplified form of \(\sqrt{48}+\sqrt{75}-\sqrt{27}\)?
B. \(6\sqrt{3}\)
\(\sqrt{48}=4\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), and \(\sqrt{27}=3\sqrt{3}\).
\(4\sqrt{3}+5\sqrt{3}-3\sqrt{3}=6\sqrt{3}\).
For like surds, work with the coefficients. चरण 1: \(\sqrt{48}=4\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), और \(\sqrt{27}=3\sqrt{3}\)। चरण 2: \(4\sqrt{3}+5\sqrt{3}-3\sqrt{3}=6\sqrt{3}\)। चरण 3: एक ही मूल वाले पदों में गुणांकों पर काम करें।
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Which option correctly states the relation between \(\sqrt{3}\) and \(\sqrt{12}\)?
B. \(\sqrt{12}=2\sqrt{3}\)
\(12=4\times3\).
\(\sqrt{12}=\sqrt{4}\sqrt{3}=2\sqrt{3}\).
Take the perfect square factor outside the radical. चरण 1: \(12=4\times3\) है। चरण 2: \(\sqrt{12}=\sqrt{4}\sqrt{3}=2\sqrt{3}\)। चरण 3: पूर्ण वर्ग गुणनखंड को मूल से बाहर निकालें।
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In which option is \(\frac{\sqrt{a}}{\sqrt{b}}\) irrational?
B. (a=50,b=2)
\(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\).
For (a=50,b=2), it becomes \(\sqrt{25}=5\), which is rational, so it should not be selected.
For an irrational quotient, \(\frac{a}{b}\) should not be a perfect square; none of the listed options gives that. चरण 1: \(\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}\) है। चरण 2: (a=50,b=2) पर \(\sqrt{\frac{50}{2}}=\sqrt{25}=5\), यह परिमेय है; इसलिए इसे नहीं चुनना चाहिए। चरण 3: सही अपरिमेय के लिए भागफल पूर्ण वर्ग न हो, जैसे यहाँ दिए विकल्पों में कोई अपरिमेय परिणाम नहीं बनता।
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Which option gives the correct value of \(\frac{\sqrt{27}}{\sqrt{3}}\)?
B. \(\sqrt{9}\) और इसलिए (3)\(\sqrt{9}\) and hence (3)
\(\frac{\sqrt{27}}{\sqrt{3}}=\sqrt{\frac{27}{3}}\).
This is \(\sqrt{9}=3\), which is rational.
In division, simplifying the radicals together is a quick method. चरण 1: \(\frac{\sqrt{27}}{\sqrt{3}}=\sqrt{\frac{27}{3}}\) लिखा जा सकता है। चरण 2: यह \(\sqrt{9}=3\) है, जो परिमेय है। चरण 3: भाग में मूलों को एक साथ सरल करना जल्दी तरीका है।
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Which option is equal to \(2\sqrt{15}\)?
B. \(\sqrt{60}\)
\(2\sqrt{15}=\sqrt{4}\sqrt{15}\).
Therefore \(2\sqrt{15}=\sqrt{60}\).
When moving a coefficient inside a square root, its square goes inside. चरण 1: \(2\sqrt{15}=\sqrt{4}\sqrt{15}\) लिखा जा सकता है। चरण 2: इसलिए \(2\sqrt{15}=\sqrt{60}\)। चरण 3: गुणांक को मूल के अंदर ले जाते समय उसका वर्ग अंदर जाता है।
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Which option gives the correct simplified form and nature of \(\sqrt{2}+\sqrt{18}\)?
A. \(4\sqrt{2}\), अपरिमेय\(4\sqrt{2}\), irrational
\(\sqrt{18}=3\sqrt{2}\).
\(\sqrt{2}+\sqrt{18}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}\), which is irrational.
For like surds, add only the outside coefficients. चरण 1: \(\sqrt{18}=3\sqrt{2}\) होता है। चरण 2: \(\sqrt{2}+\sqrt{18}=\sqrt{2}+3\sqrt{2}=4\sqrt{2}\), जो अपरिमेय है। चरण 3: समान मूल वाले पदों में केवल बाहर के गुणांक जोड़ें।
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If \(x=\sqrt{11}+\sqrt{44}\), what is the simplified form and nature of (x)?
A. \(3\sqrt{11}\), अपरिमेय\(3\sqrt{11}\), irrational
\(\sqrt{44}=\sqrt{4\times11}=2\sqrt{11}\).
Hence \(x=\sqrt{11}+2\sqrt{11}=3\sqrt{11}\), and \(\sqrt{11}\) is irrational.
For like surds, add only the coefficients, not the numbers inside the roots. चरण 1: \(\sqrt{44}=\sqrt{4\times11}=2\sqrt{11}\) होता है। चरण 2: इसलिए \(x=\sqrt{11}+2\sqrt{11}=3\sqrt{11}\), और \(\sqrt{11}\) अपरिमेय है। चरण 3: समान मूल वाले पदों में केवल गुणांक जोड़ें, मूल के अंदर की संख्या नहीं।
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Which statement is correct for \(\sqrt{12}\)?
B. यह \(2\sqrt{3}\) के बराबर है और अपरिमेय हैIt is equal to \(2\sqrt{3}\) and irrational
\(12=4\times3\).
\(\sqrt{12}=2\sqrt{3}\), and \(\sqrt{3}\) is irrational.
After simplification, if a non-square remains inside the root, the number stays irrational. चरण 1: \(12=4\times3\) है। चरण 2: \(\sqrt{12}=2\sqrt{3}\), और \(\sqrt{3}\) अपरिमेय है। चरण 3: मूल को सरल करने के बाद भी अंदर पूर्ण वर्ग न बचे तो संख्या अपरिमेय रहती है।
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What is the nature of \(\sqrt{7}+\sqrt{28}+\sqrt{63}\)?
B. अपरिमेयIrrational
\(\sqrt{28}=2\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\).
The total is \(6\sqrt{7}\), which is irrational.
Simplify an expression before deciding its nature. चरण 1: \(\sqrt{28}=2\sqrt{7}\) और \(\sqrt{63}=3\sqrt{7}\)। चरण 2: कुल योग \(6\sqrt{7}\) है, जो अपरिमेय है। चरण 3: किसी अभिव्यक्ति की प्रकृति तय करने से पहले उसे सरल करें।
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What is the nature of \(\sqrt{5}+\sqrt{20}+\sqrt{45}\)?
B. अपरिमेयIrrational
\(\sqrt{20}=2\sqrt{5}\) and \(\sqrt{45}=3\sqrt{5}\).
The total is \(6\sqrt{5}\), which is irrational.
Simplify an expression before deciding its nature. चरण 1: \(\sqrt{20}=2\sqrt{5}\) और \(\sqrt{45}=3\sqrt{5}\)। चरण 2: कुल योग \(6\sqrt{5}\) है, जो अपरिमेय है। चरण 3: किसी अभिव्यक्ति की प्रकृति तय करने से पहले उसे सरल करें।
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What is the nature of \(\sqrt{3}+\sqrt{12}+\sqrt{27}\)?
B. अपरिमेयIrrational
\(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\).
The total is \(6\sqrt{3}\), which is irrational.
Decide the nature of the number only after simplification. चरण 1: \(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{27}=3\sqrt{3}\)। चरण 2: कुल योग \(6\sqrt{3}\) है, जो अपरिमेय है। चरण 3: सरलीकरण के बाद ही संख्या की प्रकृति तय करें।
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Which of the following results is rational?
B. \(\sqrt{2}\times\sqrt{8}\)
\(\sqrt{2}\times\sqrt{8}=\sqrt{16}\).
\(\sqrt{16}=4\), which is rational.
The product of two irrational numbers is not always irrational. चरण 1: \(\sqrt{2}\times\sqrt{8}=\sqrt{16}\) होता है। चरण 2: \(\sqrt{16}=4\), जो परिमेय संख्या है। चरण 3: दो अपरिमेय संख्याओं का गुणनफल हमेशा अपरिमेय नहीं होता।
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Which of the following results is irrational?
B. \(\sqrt{7}+\sqrt{28}\)
\(\sqrt{28}=2\sqrt{7}\), so \(\sqrt{7}+\sqrt{28}=3\sqrt{7}\).
\(3\sqrt{7}\) is irrational because \(\sqrt{7}\) is irrational.
In options, simplify first before deciding the nature of the result. चरण 1: \(\sqrt{28}=2\sqrt{7}\), इसलिए \(\sqrt{7}+\sqrt{28}=3\sqrt{7}\)। चरण 2: \(3\sqrt{7}\) अपरिमेय है क्योंकि \(\sqrt{7}\) अपरिमेय है। चरण 3: विकल्पों में परिणाम की प्रकृति तय करने से पहले सरलीकरण करें।
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What is \(\sqrt{7}+2\sqrt{7}\) equal to?
B. \(3\sqrt{7}\)
Both terms contain the same radical \(\sqrt{7}\).
\(1\sqrt{7}+2\sqrt{7}=3\sqrt{7}\).
For like radicals, add only the outside coefficients. चरण 1: दोनों पदों में \(\sqrt{7}\) समान है। चरण 2: \(1\sqrt{7}+2\sqrt{7}=3\sqrt{7}\)। चरण 3: समान वर्गमूलों में केवल बाहर के गुणांक जोड़ें।
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What is the simplified form of \(\sqrt{10}+\sqrt{10}+\sqrt{10}+\sqrt{10}\)?
A. \(4\sqrt{10}\)
Four like radical terms are being added.
\(\sqrt{10}+\sqrt{10}+\sqrt{10}+\sqrt{10}=4\sqrt{10}\).
When adding like radicals, add only the coefficients. चरण 1: चार समान वर्गमूल पद जोड़े जा रहे हैं। चरण 2: \(\sqrt{10}+\sqrt{10}+\sqrt{10}+\sqrt{10}=4\sqrt{10}\)। चरण 3: समान वर्गमूलों को जोड़ते समय केवल गुणांक जोड़ें।
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What will be the simplified form of \(\sqrt{11}+\sqrt{11}\)?
C. \(2\sqrt{11}\)
Both terms have the same square root.
\(\sqrt{11}+\sqrt{11}=2\sqrt{11}\).
For like radicals, add only the coefficients, not the numbers inside the roots. चरण 1: दोनों पद समान वर्गमूल वाले हैं। चरण 2: \(\sqrt{11}+\sqrt{11}=2\sqrt{11}\)। चरण 3: समान वर्गमूलों में अंदर की संख्याएँ नहीं, केवल गुणांक जोड़े जाते हैं।
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Which of the following results is irrational?
A. \(\sqrt{2}+\sqrt{8}\)
\(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\).
\(3\sqrt{2}\) is irrational.
In options, simplify the result before deciding its nature. चरण 1: \(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\)। चरण 2: \(3\sqrt{2}\) अपरिमेय है। चरण 3: विकल्पों में परिणाम निकालकर ही संख्या की प्रकृति तय करें।
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What is \(\sqrt{3}+2\sqrt{3}\) equal to?
A. \(3\sqrt{3}\)
Both terms have the same radical \(\sqrt{3}\).
\(1\sqrt{3}+2\sqrt{3}=3\sqrt{3}\).
For like radicals, add only the coefficients. चरण 1: दोनों पदों में \(\sqrt{3}\) समान है। चरण 2: \(1\sqrt{3}+2\sqrt{3}=3\sqrt{3}\)। चरण 3: समान वर्गमूलों में केवल गुणांक जोड़ें।
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What is the simplified form of \(\sqrt{6}+\sqrt{6}+\sqrt{6}\)?
A. \(3\sqrt{6}\)
Three like irrational terms are being added.
\(\sqrt{6}+\sqrt{6}+\sqrt{6}=3\sqrt{6}\).
Count like radicals as coefficients. चरण 1: तीन समान अपरिमेय पद जोड़े जा रहे हैं। चरण 2: \(\sqrt{6}+\sqrt{6}+\sqrt{6}=3\sqrt{6}\)। चरण 3: समान वर्गमूलों को गुणांक की तरह गिनें।
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What do we get after simplifying \(\sqrt{48}\)?
A. \(4\sqrt{3}\)
\(48=16 \times 3\).
\(\sqrt{48}=\sqrt{16 \times 3}=4\sqrt{3}\).
Using the largest perfect square gives the simplest form. चरण 1: \(48=16 \times 3\) है। चरण 2: \(\sqrt{48}=\sqrt{16 \times 3}=4\sqrt{3}\)। चरण 3: सबसे बड़ा पूर्ण वर्ग लेने से सरल रूप सही और छोटा मिलता है।
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What is the simplified form of \(\sqrt{75}\)?
A. \(5\sqrt{3}\)
Write \(75=25 \times 3\).
\(\sqrt{75}=\sqrt{25 \times 3}=5\sqrt{3}\).
While simplifying a square root, split the inside number into a perfect square and the remaining factor. चरण 1: \(75=25 \times 3\) लिखें। चरण 2: \(\sqrt{75}=\sqrt{25 \times 3}=5\sqrt{3}\)। चरण 3: वर्गमूल को सरल करते समय अंदर की संख्या को पूर्ण वर्ग और बाकी गुणनखंड में तोड़ें।
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Which statement is correct about \(\sqrt{2}+\sqrt{3}\)?
B. यह अपरिमेय हैIt is irrational
Both \(\sqrt{2}\) and \(\sqrt{3}\) are irrational.
Their sum is not \(\sqrt{5}\); \(\sqrt{2}+\sqrt{3}\) remains irrational.
Do not add the numbers inside different square roots directly. चरण 1: \(\sqrt{2}\) और \(\sqrt{3}\) दोनों अपरिमेय हैं। चरण 2: उनका योग \(\sqrt{5}\) नहीं होता; \(\sqrt{2}+\sqrt{3}\) अपरिमेय रहता है। चरण 3: अलग-अलग वर्गमूलों को सीधे अंदर की संख्याएँ जोड़कर न लिखें।
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What is the simplified form of \(\sqrt{27}\)?
A. \(3\sqrt{3}\)
Write \(27=9 \times 3\).
\(\sqrt{27}=\sqrt{9 \times 3}=3\sqrt{3}\).
When (9) appears as a factor inside a square root, take it out as (3). चरण 1: \(27=9 \times 3\) लिखें। चरण 2: \(\sqrt{27}=\sqrt{9 \times 3}=3\sqrt{3}\)। चरण 3: वर्गमूल में पूर्ण वर्ग (9) दिखे तो उसे बाहर (3) के रूप में निकालें।
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The simplified form of \(\sqrt{2}+\sqrt{2}\) is what type of number?
B. अपरिमेय संख्याIrrational number
\(\sqrt{2}+\sqrt{2}=2\sqrt{2}\).
\(\sqrt{2}\) is irrational and (2) is non-zero rational, so \(2\sqrt{2}\) is irrational.
First simplify like radicals, then decide the type. चरण 1: \(\sqrt{2}+\sqrt{2}=2\sqrt{2}\)। चरण 2: \(\sqrt{2}\) अपरिमेय है और (2) अशून्य परिमेय है, इसलिए \(2\sqrt{2}\) अपरिमेय है। चरण 3: समान वर्गमूलों को पहले जोड़कर सरल करें।
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