25 results found for "conjugate multiplication" in Class 10.
Question
Expert Mathematics
Polynomials Irrational numbers and real numbers Class 10 Level 27
कौन सा विकल्प (\(\sqrt{14}+\sqrt{6}\)\(\sqrt{14}-\sqrt{6}\)+\sqrt{84}) के बराबर है?
Which option is equal to (\(\sqrt{14}+\sqrt{6}\)\(\sqrt{14}-\sqrt{6}\)+\sqrt{84})?
#conjugates
#surd-simplification
#expression
A \(8+2\sqrt{21}\)
B \(20+2\sqrt{21}\)
C \(8+\sqrt{21}\)
D \(20+\sqrt{84}\)
Explanation opens after your attempt
Correct Answer
A. \(8+2\sqrt{21}\)
Step 1
Concept
The first product is (14-6=8), and \(\sqrt{84}=2\sqrt{21}\). In exams use both conjugate multiplication and radical simplification.
Step 2
Why this answer is correct
The correct answer is A. \(8+2\sqrt{21}\). The first product is (14-6=8), and \(\sqrt{84}=2\sqrt{21}\). In exams use both conjugate multiplication and radical simplification.
Step 3
Exam Tip
पहला गुणनफल (14-6=8) है और \(\sqrt{84}=2\sqrt{21}\) है। परीक्षा में संयुग्मी गुणन और मूल सरलीकरण दोनों करें।
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Question
Expert Mathematics
Polynomials Irrational numbers and real numbers Class 10 Level 27
यदि \(\alpha=5+2\sqrt{6}\) और \(\beta=5-2\sqrt{6}\), तो \(\alpha\beta\) क्या है?
If \(\alpha=5+2\sqrt{6}\) and \(\beta=5-2\sqrt{6}\), what is \(\alpha\beta\)?
#conjugates
#product
#irrational-numbers
A (1)
B (25)
C (24)
D (10)
Explanation opens after your attempt
Step 1
Concept
(\alpha\beta=25-\(2\sqrt{6}\)2 =25-24=1). In exams square terms correctly in conjugate multiplication.
Step 2
Why this answer is correct
The correct answer is A. (1). (\alpha\beta=25-\(2\sqrt{6}\)2 =25-24=1). In exams square terms correctly in conjugate multiplication.
Step 3
Exam Tip
(\alpha\beta=25-\(2\sqrt{6}\)2 =25-24=1) है। परीक्षा में संयुग्मी गुणन में वर्ग सही करें।
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Question
Expert Mathematics
Polynomials Irrational numbers and real numbers Class 10 Level 27
किस विकल्प में दो अपरिमेय संख्याओं का गुणनफल परिमेय है?
In which option is the product of two irrational numbers rational?
#irrational-product
#counterexample
#conjugates
A (\(2+\sqrt{3}\)\(2-\sqrt{3}\))
B (\(\sqrt{2}\)\(\sqrt{5}\))
C (\(\sqrt{3}\)\(\sqrt{7}\))
D (\(\sqrt{5}\)\(\sqrt{6}\))
Explanation opens after your attempt
Correct Answer
A. (\(2+\sqrt{3}\)\(2-\sqrt{3}\))
Step 1
Concept
(\(2+\sqrt{3}\)\(2-\sqrt{3}\)=4-3=1) which is rational. In exams remember conjugate multiplication as a counterexample.
Step 2
Why this answer is correct
The correct answer is A. (\(2+\sqrt{3}\)\(2-\sqrt{3}\)). (\(2+\sqrt{3}\)\(2-\sqrt{3}\)=4-3=1) which is rational. In exams remember conjugate multiplication as a counterexample.
Step 3
Exam Tip
(\(2+\sqrt{3}\)\(2-\sqrt{3}\)=4-3=1) है जो परिमेय है। परीक्षा में संयुग्मी गुणन को प्रतिउदाहरण के रूप में याद रखें।
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Question
Expert Mathematics
Polynomials Irrational numbers and real numbers Class 10 Level 27
यदि \(a=\sqrt{13}+\sqrt{6}\) और \(b=\sqrt{13}-\sqrt{6}\), तो (ab) का मान क्या है?
If \(a=\sqrt{13}+\sqrt{6}\) and \(b=\sqrt{13}-\sqrt{6}\), what is the value of (ab)?
#conjugates
#surd-product
#rational-number
A (7)
B (19)
C \(\sqrt{78}\)
D \(2\sqrt{78}\)
Explanation opens after your attempt
Step 1
Concept
(ab=13-6=7). In exams conjugate multiplication removes radicals.
Step 2
Why this answer is correct
The correct answer is A. (7). (ab=13-6=7). In exams conjugate multiplication removes radicals.
Step 3
Exam Tip
(ab=13-6=7) है। परीक्षा में संयुग्मी गुणन से मूल हट जाते हैं।
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Question
Expert Mathematics
Polynomials Irrational numbers and real numbers Class 10 Level 26
यदि \(a=\sqrt{11}+\sqrt{5}\) और \(b=\sqrt{11}-\sqrt{5}\), तो (ab) क्या होगा?
If \(a=\sqrt{11}+\sqrt{5}\) and \(b=\sqrt{11}-\sqrt{5}\), what is (ab)?
#conjugates
#real-numbers
#surds
A (6)
B \(\sqrt{55}\)
C (16)
D \(2\sqrt{55}\)
Explanation opens after your attempt
Step 1
Concept
(ab=\(\sqrt{11}\)2 -\(\sqrt{5}\)2 =11-5=6). In exams conjugate multiplication removes radicals.
Step 2
Why this answer is correct
The correct answer is A. (6). (ab=\(\sqrt{11}\)2 -\(\sqrt{5}\)2 =11-5=6). In exams conjugate multiplication removes radicals.
Step 3
Exam Tip
(ab=\(\sqrt{11}\)2 -\(\sqrt{5}\)2 =11-5=6) है। परीक्षा में संयुग्मी गुणन से मूल हट जाते हैं।
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Question
Hard Mathematics
Polynomials Irrational numbers and real numbers Class 10 Level 26
यदि \(x=\sqrt{6}+\sqrt{2}\) और \(y=\sqrt{6}-\sqrt{2}\), तो (xy) क्या है?
If \(x=\sqrt{6}+\sqrt{2}\) and \(y=\sqrt{6}-\sqrt{2}\), what is (xy)?
#conjugate multiplication
#surds
#real numbers
A (4)
B (8)
C \(2\sqrt{12}\)
D \(6+\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
(xy=\(\sqrt{6}\)2 -\(\sqrt{2}\)2 =6-2=4). Conjugate multiplication saves time in exams.
Step 2
Why this answer is correct
The correct answer is A. (4). (xy=\(\sqrt{6}\)2 -\(\sqrt{2}\)2 =6-2=4). Conjugate multiplication saves time in exams.
Step 3
Exam Tip
(xy=\(\sqrt{6}\)2 -\(\sqrt{2}\)2 =6-2=4) है। परीक्षा में संयुग्मी गुणन से समय बचता है।
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Question
Hard Mathematics
Polynomials Irrational numbers and real numbers Class 10 Level 28
यदि \(\alpha=\sqrt{5}+2\) और \(\beta=\sqrt{5}-2\), तो \(\alpha\beta\) क्या है?
If \(\alpha=\sqrt{5}+2\) and \(\beta=\sqrt{5}-2\), what is \(\alpha\beta\)?
#conjugates
#product
#real numbers
A (1)
B (9)
C \(\sqrt{5}\)
D \(4\sqrt{5}\)
Explanation opens after your attempt
Step 1
Concept
(\alpha\beta=\(\sqrt{5}+2\)\(\sqrt{5}-2\)=5-4=1). Conjugate multiplication often gives a rational answer.
Step 2
Why this answer is correct
The correct answer is A. (1). (\alpha\beta=\(\sqrt{5}+2\)\(\sqrt{5}-2\)=5-4=1). Conjugate multiplication often gives a rational answer.
Step 3
Exam Tip
(\alpha\beta=\(\sqrt{5}+2\)\(\sqrt{5}-2\)=5-4=1) है। परीक्षा में संयुग्मी गुणन से परिमेय उत्तर मिलता है।
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Question
Hard Mathematics
Polynomials Irrational numbers and real numbers Class 10 Level 28
यदि \(\alpha=\sqrt{3}+1\) और \(\beta=\sqrt{3}-1\), तो \(\alpha\beta\) क्या है?
If \(\alpha=\sqrt{3}+1\) and \(\beta=\sqrt{3}-1\), what is \(\alpha\beta\)?
#conjugate
#product
#irrational-numbers
A (2)
B (4)
C \(\sqrt{3}\)
D \(3+\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
(\alpha\beta=\(\sqrt{3}+1\)\(\sqrt{3}-1\)=3-1=2). The irrational part cancels in conjugate multiplication.
Step 2
Why this answer is correct
The correct answer is A. (2). (\alpha\beta=\(\sqrt{3}+1\)\(\sqrt{3}-1\)=3-1=2). The irrational part cancels in conjugate multiplication.
Step 3
Exam Tip
(\alpha\beta=\(\sqrt{3}+1\)\(\sqrt{3}-1\)=3-1=2) है। संयुग्मी गुणनफल से अपरिमेय भाग कट जाता है।
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Question
Hard Mathematics
Polynomials Irrational numbers and real numbers Class 10 Level 25
यदि \(a=\sqrt{7}+\sqrt{2}\) और \(b=\sqrt{7}-\sqrt{2}\) हैं तो (ab) का मान क्या है?
If \(a=\sqrt{7}+\sqrt{2}\) and \(b=\sqrt{7}-\sqrt{2}\), what is the value of (ab)?
#conjugate
#product
#rational-result
A (5)
B (9)
C \(\sqrt{14}\)
D \(7+2\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
In conjugate multiplication, (ab=7-2=5). The difference of squares formula gives the answer quickly.
Step 2
Why this answer is correct
The correct answer is A. (5). In conjugate multiplication, (ab=7-2=5). The difference of squares formula gives the answer quickly.
Step 3
Exam Tip
संयुग्मी गुणन में (ab=7-2=5) होता है। अंतर वर्ग सूत्र जल्दी उत्तर देता है।
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Question
Medium Mathematics
Polynomials Irrational numbers and real numbers Class 10 Level 27
यदि \(p=7+\sqrt{11}\) और \(q=7-\sqrt{11}\) हैं तो (pq) का मान क्या है?
If \(p=7+\sqrt{11}\) and \(q=7-\sqrt{11}\), what is the value of (pq)?
#conjugate
#surds
#product
A (38)
B (60)
C \(14\sqrt{11}\)
D \(49+\sqrt{11}\)
Explanation opens after your attempt
Step 1
Concept
Conjugate multiplication gives (pq=72 -\(\sqrt{11}\)2 =49-11=38). Use \(a^2-b^2\) in such questions.
Step 2
Why this answer is correct
The correct answer is A. (38). Conjugate multiplication gives (pq=72 -\(\sqrt{11}\)2 =49-11=38). Use \(a^2-b^2\) in such questions.
Step 3
Exam Tip
संयुग्मी गुणन से (pq=72 -\(\sqrt{11}\)2 =49-11=38) मिलता है। ऐसे प्रश्नों में \(a^2-b^2\) लगाएँ।
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Question
Medium Mathematics
Polynomials Irrational numbers and real numbers Class 10 Level 27
कौन सा विकल्प (\(\sqrt{15}+\sqrt{6}\)\(\sqrt{15}-\sqrt{6}\)) का मान है?
Which option is the value of (\(\sqrt{15}+\sqrt{6}\)\(\sqrt{15}-\sqrt{6}\))?
#conjugate
#identity
#rational-result
A (9)
B (21)
C \(\sqrt{90}\)
D \(15+6\sqrt{6}\)
Explanation opens after your attempt
Step 1
Concept
This is the difference of squares formula and the value is (15-6=9). In conjugate multiplication the irrational part cancels.
Step 2
Why this answer is correct
The correct answer is A. (9). This is the difference of squares formula and the value is (15-6=9). In conjugate multiplication the irrational part cancels.
Step 3
Exam Tip
यह अंतर वर्ग सूत्र है और मान (15-6=9) मिलता है। संयुग्मी गुणन में अपरिमेय भाग हट जाता है।
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Question
Medium Mathematics
Polynomials Irrational numbers and real numbers Class 10 Level 27
कौन सा विकल्प \(2+\sqrt{5}\) और \(2-\sqrt{5}\) के गुणनफल का मान है?
Which option is the value of the product of \(2+\sqrt{5}\) and \(2-\sqrt{5}\)?
#conjugate
#surds
#rational-result
A (-1)
B (1)
C \(4+\sqrt{5}\)
D \(4-\sqrt{5}\)
Explanation opens after your attempt
Step 1
Concept
(\(2+\sqrt{5}\)\(2-\sqrt{5}\)=4-5=-1). In conjugate multiplication the irrational part cancels.
Step 2
Why this answer is correct
The correct answer is A. (-1). (\(2+\sqrt{5}\)\(2-\sqrt{5}\)=4-5=-1). In conjugate multiplication the irrational part cancels.
Step 3
Exam Tip
(\(2+\sqrt{5}\)\(2-\sqrt{5}\)=4-5=-1) है। संयुग्मी गुणन में अपरिमेय भाग हट जाता है।
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Question
Medium Mathematics
Polynomials Irrational numbers and real numbers Class 10 Level 26
कौन सा विकल्प (\(4+\sqrt{7}\)\(4-\sqrt{7}\)) का मान है?
Which option is the value of (\(4+\sqrt{7}\)\(4-\sqrt{7}\))?
#conjugate
#surds
#rational-result
A (9)
B (23)
C (16+7)
D \(8\sqrt{7}\)
Explanation opens after your attempt
Step 1
Concept
Conjugate multiplication gives (42 -\(\sqrt{7}\)2 =16-7=9). Use the difference of squares formula in such questions.
Step 2
Why this answer is correct
The correct answer is A. (9). Conjugate multiplication gives (42 -\(\sqrt{7}\)2 =16-7=9). Use the difference of squares formula in such questions.
Step 3
Exam Tip
संयुग्मी गुणन से (42 -\(\sqrt{7}\)2 =16-7=9) मिलता है। ऐसे प्रश्नों में अंतर वर्ग सूत्र लगाएँ।
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Question
Medium Mathematics
Polynomials Irrational numbers and real numbers Class 10 Level 26
कौन सा विकल्प \(1+\sqrt{7}\) और \(1-\sqrt{7}\) के गुणनफल का मान है?
Which option is the value of the product of \(1+\sqrt{7}\) and \(1-\sqrt{7}\)?
#conjugate
#surds
#rational-result
A (-6)
B (6)
C \(1+\sqrt{7}\)
D (7)
Explanation opens after your attempt
Step 1
Concept
(\(1+\sqrt{7}\)\(1-\sqrt{7}\)=1-7=-6). In conjugate multiplication the irrational part cancels.
Step 2
Why this answer is correct
The correct answer is A. (-6). (\(1+\sqrt{7}\)\(1-\sqrt{7}\)=1-7=-6). In conjugate multiplication the irrational part cancels.
Step 3
Exam Tip
(\(1+\sqrt{7}\)\(1-\sqrt{7}\)=1-7=-6) है। संयुग्मी गुणन में अपरिमेय भाग हट जाता है।
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Question
Medium Mathematics
Polynomials Irrational numbers and real numbers Class 10 Level 26
यदि \(u=6+\sqrt{5}\) और \(v=6-\sqrt{5}\) हैं तो (uv) का मान क्या है?
If \(u=6+\sqrt{5}\) and \(v=6-\sqrt{5}\), what is the value of (uv)?
#conjugate
#surds
#rational-result
A (31)
B \(36+\sqrt{5}\)
C \(12\sqrt{5}\)
D (41)
Explanation opens after your attempt
Step 1
Concept
Conjugate multiplication gives ((6)2 -\(\sqrt{5}\)2 =36-5=31). Use \(a^2-b^2\) in such questions.
Step 2
Why this answer is correct
The correct answer is A. (31). Conjugate multiplication gives ((6)2 -\(\sqrt{5}\)2 =36-5=31). Use \(a^2-b^2\) in such questions.
Step 3
Exam Tip
संयुग्मी गुणन से ((6)2 -\(\sqrt{5}\)2 =36-5=31) मिलता है। ऐसे प्रश्नों में \(a^2-b^2\) प्रयोग करें।
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Question
Medium Mathematics
Polynomials Irrational numbers and real numbers Class 10 Level 25
यदि \(u=5+\sqrt{2}\) और \(v=5-\sqrt{2}\) हैं तो (uv) का मान क्या है?
If \(u=5+\sqrt{2}\) and \(v=5-\sqrt{2}\), what is the value of (uv)?
#conjugate
#surds
#rational-result
A (23)
B (27)
C \(25+\sqrt{2}\)
D \(10\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
(\(5+\sqrt{2}\)\(5-\sqrt{2}\)=25-2=23). In conjugate multiplication the irrational part cancels.
Step 2
Why this answer is correct
The correct answer is A. (23). (\(5+\sqrt{2}\)\(5-\sqrt{2}\)=25-2=23). In conjugate multiplication the irrational part cancels.
Step 3
Exam Tip
(\(5+\sqrt{2}\)\(5-\sqrt{2}\)=25-2=23) है। संयुग्मी गुणन में अपरिमेय भाग हट जाता है।
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Question
Expert Mathematics
Real Numbers 5: Irrational numbers Class 10 Level 13
यदि \(a=\sqrt{8}+\sqrt{18}\) और \(b=\sqrt{8}-\sqrt{18}\), तो (ab) का मान क्या है?
If \(a=\sqrt{8}+\sqrt{18}\) and \(b=\sqrt{8}-\sqrt{18}\), what is the value of (ab)?
#conjugate product
#negative rational
#class 10
A (-10)
B (10)
C (26)
D \(12\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
(ab=\(\sqrt{8}\)2 -\(\sqrt{18}\)2 ).
Step 2
Why this answer is correct
(ab=8-18=-10), which is rational.
Step 3
Exam Tip
In conjugate multiplication, you do not always need to simplify each radical first. चरण 1: (ab=\(\sqrt{8}\)2 -\(\sqrt{18}\)2 ) है। चरण 2: (ab=8-18=-10), जो परिमेय है। चरण 3: संयुग्मी गुणन में मूलों को अलग-अलग सरल करना जरूरी नहीं होता।
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Question
Hard Mathematics
Real Numbers 5: Irrational numbers Class 10 Level 15
यदि \(a=5+\sqrt{7}\) और \(b=5-\sqrt{7}\), तो (ab) का मान क्या है?
If \(a=5+\sqrt{7}\) and \(b=5-\sqrt{7}\), what is the value of (ab)?
#conjugate surds
#rational product
#class 10
A (18)
B \(25+\sqrt{7}\)
C (32)
D \(10\sqrt{7}\)
Explanation opens after your attempt
Step 1
Concept
This is multiplication of conjugates.
Step 2
Why this answer is correct
(ab=52 -\(\sqrt{7}\)2 =25-7=18).
Step 3
Exam Tip
In conjugate multiplication, the middle irrational terms cancel. चरण 1: यह संयुग्मी संख्याओं का गुणन है। चरण 2: (ab=52 -\(\sqrt{7}\)2 =25-7=18)। चरण 3: संयुग्मी गुणन में बीच के अपरिमेय पद कट जाते हैं।
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Question
Hard Mathematics
Real Numbers 5: Irrational numbers Class 10 Level 14
यदि \(a=\sqrt{3}+2\) और \(b=\sqrt{3}-2\), तो (ab) की प्रकृति क्या है?
If \(a=\sqrt{3}+2\) and \(b=\sqrt{3}-2\), what is the nature of (ab)?
#conjugate surds
#rational product
#class 10
#hard
A परिमेय और ऋणात्मक / Rational and negative
B अपरिमेय और धनात्मक / Irrational and positive
C परिमेय और धनात्मक / Rational and positive
D अपरिमेय और ऋणात्मक / Irrational and negative
Explanation opens after your attempt
Correct Answer
A. परिमेय और ऋणात्मक / Rational and negative
Step 1
Concept
(a) and (b) are conjugates.
Step 2
Why this answer is correct
(ab=\(\sqrt{3}\)2 -22 =3-4=-1), which is rational and negative.
Step 3
Exam Tip
In conjugate multiplication, the middle irrational terms cancel. चरण 1: (a) और (b) संयुग्मी रूप में हैं। चरण 2: (ab=\(\sqrt{3}\)2 -22 =3-4=-1), जो परिमेय और ऋणात्मक है। चरण 3: संयुग्मी गुणन में बीच के अपरिमेय पद कट जाते हैं।
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Question
Hard Mathematics
Real Numbers 5: Irrational numbers Class 10 Level 14
निम्न में से कौन-सा मान \(1+\sqrt{2}\) और \(1-\sqrt{2}\) के गुणनफल के बराबर है?
Which value equals the product of \(1+\sqrt{2}\) and \(1-\sqrt{2}\)?
#conjugates
#difference of squares
#class 10
A (1)
B (3)
C (-1)
D \(2\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
This is a product of conjugates.
Step 2
Why this answer is correct
(\(1+\sqrt{2}\)\(1-\sqrt{2}\)=1-\(\sqrt{2}\)2 =1-2=-1).
Step 3
Exam Tip
In conjugate multiplication, the middle irrational terms cancel. चरण 1: यह संयुग्मी संख्याओं का गुणन है। चरण 2: (\(1+\sqrt{2}\)\(1-\sqrt{2}\)=1-\(\sqrt{2}\)2 =1-2=-1)। चरण 3: संयुग्मी गुणन में बीच के अपरिमेय पद कट जाते हैं।
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Question
Medium Mathematics
Real Numbers 5: Irrational numbers Class 10 Level 15
(\left\(\sqrt{13}-\sqrt{5}\right\)\left\(\sqrt{13}+\sqrt{5}\right\)) का मान क्या है?
What is the value of (\left\(\sqrt{13}-\sqrt{5}\right\)\left\(\sqrt{13}+\sqrt{5}\right\))?
#real-numbers
#conjugate-product
#rational-result
A (8)
B (18)
C \(\sqrt{65}\)
D \(13+\sqrt{5}\)
Explanation opens after your attempt
Step 1
Concept
This is of the form ((a-b)(a+b)=a-2 -b-2 ).
Step 2
Why this answer is correct
(\(\sqrt{13}\)2 -\(\sqrt{5}\)2 =13-5=8).
Step 3
Exam Tip
In conjugate multiplication, directly use the difference of squares. चरण 1: यह ((a-b)(a+b)=a-2 -b-2 ) का रूप है। चरण 2: (\(\sqrt{13}\)2 -\(\sqrt{5}\)2 =13-5=8)। चरण 3: संयुग्म गुणन में वर्गों का अंतर सीधे लगाएं।
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Question
Medium Mathematics
Real Numbers 5: Irrational numbers Class 10 Level 15
(\left\(4+\sqrt{7}\right\)\left\(4-\sqrt{7}\right\)) का मान क्या है?
What is the value of (\left\(4+\sqrt{7}\right\)\left\(4-\sqrt{7}\right\))?
#real-numbers
#conjugates
#rational-result
A (9)
B (23)
C \(16+\sqrt{7}\)
D \(16-\sqrt{7}\)
Explanation opens after your attempt
Step 1
Concept
This is of the form ((a+b)(a-b)=a-2 -b-2 ).
Step 2
Why this answer is correct
(42 -\(\sqrt{7}\)2 =16-7=9).
Step 3
Exam Tip
In conjugate multiplication, directly use the difference of squares. चरण 1: यह ((a+b)(a-b)=a-2 -b-2 ) का रूप है। चरण 2: (42 -\(\sqrt{7}\)2 =16-7=9)। चरण 3: संयुग्म गुणन में वर्गों का अंतर सीधे लगाएं।
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Question
Medium Mathematics
Real Numbers 5: Irrational numbers Class 10 Level 14
(\left\(\sqrt{11}-\sqrt{2}\right\)\left\(\sqrt{11}+\sqrt{2}\right\)) का मान क्या है?
What is the value of (\left\(\sqrt{11}-\sqrt{2}\right\)\left\(\sqrt{11}+\sqrt{2}\right\))?
#real-numbers
#conjugate-product
#rational-result
A (9)
B (13)
C \(\sqrt{22}\)
D \(11+\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
This is of the form ((a-b)(a+b)=a-2 -b-2 ).
Step 2
Why this answer is correct
(\(\sqrt{11}\)2 -\(\sqrt{2}\)2 =11-2=9).
Step 3
Exam Tip
In conjugate multiplication, directly use the difference of squares. चरण 1: यह ((a-b)(a+b)=a-2 -b-2 ) का रूप है। चरण 2: (\(\sqrt{11}\)2 -\(\sqrt{2}\)2 =11-2=9)। चरण 3: संयुग्म गुणन में वर्गों का अंतर सीधे लगाएं।
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Question
Medium Mathematics
Real Numbers 5: Irrational numbers Class 10 Level 14
(\left\(3+\sqrt{5}\right\)\left\(3-\sqrt{5}\right\)) का मान क्या है?
What is the value of (\left\(3+\sqrt{5}\right\)\left\(3-\sqrt{5}\right\))?
#real-numbers
#conjugates
#rational-result
A (4)
B (14)
C \(9+\sqrt{5}\)
D \(9-\sqrt{5}\)
Explanation opens after your attempt
Step 1
Concept
This is of the form ((a+b)(a-b)=a-2 -b-2 ).
Step 2
Why this answer is correct
(32 -\(\sqrt{5}\)2 =9-5=4).
Step 3
Exam Tip
In conjugate multiplication, directly use difference of squares. चरण 1: यह ((a+b)(a-b)=a-2 -b-2 ) का रूप है। चरण 2: (32 -\(\sqrt{5}\)2 =9-5=4)। चरण 3: संयुग्म गुणन में वर्गों का अंतर सीधे लगाएं।
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Question
Medium Mathematics
Real Numbers 5: Irrational numbers Class 10 Level 13
(\left\(\sqrt{7}-\sqrt{3}\right\)\left\(\sqrt{7}+\sqrt{3}\right\)) का मान क्या है?
What is the value of (\left\(\sqrt{7}-\sqrt{3}\right\)\left\(\sqrt{7}+\sqrt{3}\right\))?
#real-numbers
#conjugate-product
#rational-result
A (4)
B (10)
C \(\sqrt{21}\)
D \(7+\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
This is of the form ((a-b)(a+b)=a-2 -b-2 ).
Step 2
Why this answer is correct
(\(\sqrt{7}\)2 -\(\sqrt{3}\)2 =7-3=4).
Step 3
Exam Tip
In conjugate multiplication, directly use the difference of squares. चरण 1: यह ((a-b)(a+b)=a-2 -b-2 ) का रूप है। चरण 2: (\(\sqrt{7}\)2 -\(\sqrt{3}\)2 =7-3=4)। चरण 3: संयुग्म गुणन में सीधे वर्गों का अंतर लगाएं।
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