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A. वे \(a+\sqrt{7}\) और \(a-\sqrt{7}\) हैं/They are \(a+\sqrt{7}\) and \(a-\sqrt{7}\)
Step 1
Concept
(p(x)=(x-a)2-7), so \(x=a\pm\sqrt{7}\). Recognizing a perfect-square form saves time in hard questions.
Step 2
Why this answer is correct
The correct answer is A. वे \(a+\sqrt{7}\) और \(a-\sqrt{7}\) हैं / They are \(a+\sqrt{7}\) and \(a-\sqrt{7}\). (p(x)=(x-a)2-7), so \(x=a\pm\sqrt{7}\). Recognizing a perfect-square form saves time in hard questions.
Step 3
Exam Tip
(p(x)=(x-a)2-7), इसलिए \(x=a\pm\sqrt{7}\) है। पूर्ण वर्ग रूप पहचानना कठिन प्रश्नों में समय बचाता है।
The constant term is the product, and (\(4+\sqrt{11}\)\(4-\sqrt{11}\)=16-11=5). In conjugate products, the irrational middle part cancels.
Step 2
Why this answer is correct
The correct answer is A. (5). The constant term is the product, and (\(4+\sqrt{11}\)\(4-\sqrt{11}\)=16-11=5). In conjugate products, the irrational middle part cancels.
Step 3
Exam Tip
स्थिर पद गुणनफल है और (\(4+\sqrt{11}\)\(4-\sqrt{11}\)=16-11=5)। संयुग्मी गुणनफल में बीच का अपरिमेय भाग हट जाता है।
\(\alpha+\beta=10\) and \(\alpha\beta=25-6=19\), so \(\alpha^2+\beta^2=100-38=62\). Use (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta).
Step 2
Why this answer is correct
The correct answer is A. (62). \(\alpha+\beta=10\) and \(\alpha\beta=25-6=19\), so \(\alpha^2+\beta^2=100-38=62\). Use (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta).
Step 3
Exam Tip
\(\alpha+\beta=10\) और \(\alpha\beta=25-6=19\), इसलिए \(\alpha^2+\beta^2=100-38=62\)। पहचान (\alpha-2+\beta-2=\(\alpha+\beta\)2-2\alpha\beta) उपयोग करें।
The zeroes are \(5\pm2\sqrt{2}\), so the difference is \(4\sqrt{2}\). For conjugate zeroes, the difference is twice the radical part.
Step 2
Why this answer is correct
The correct answer is A. \(4\sqrt{2}\). The zeroes are \(5\pm2\sqrt{2}\), so the difference is \(4\sqrt{2}\). For conjugate zeroes, the difference is twice the radical part.
Step 3
Exam Tip
शून्यक \(5\pm2\sqrt{2}\) हैं, इसलिए अंतर \(4\sqrt{2}\) है। संयुग्मी शून्यकों में अंतर मूल भाग का दोगुना होता है।
After removing the common factor, we get \(x^2-6x+7\), and (D=36-28=8). Since (D) is positive and not a perfect square, the zeroes are real irrational.
Step 2
Why this answer is correct
The correct answer is B. वास्तविक और अपरिमेय / Real and irrational. After removing the common factor, we get \(x^2-6x+7\), and (D=36-28=8). Since (D) is positive and not a perfect square, the zeroes are real irrational.
Step 3
Exam Tip
सामान्य गुणनखंड हटाने पर \(x^2-6x+7\) मिलता है और (D=36-28=8)। (D) धनात्मक अपूर्ण वर्ग है, इसलिए शून्यक वास्तविक अपरिमेय हैं।
\(\sqrt{12}=2\sqrt{3}\), so the sum is \(3\sqrt{3}\). In a monic polynomial, the coefficient of (x) is the negative of the sum of zeroes.
Step 2
Why this answer is correct
The correct answer is A. \(-3\sqrt{3}\). \(\sqrt{12}=2\sqrt{3}\), so the sum is \(3\sqrt{3}\). In a monic polynomial, the coefficient of (x) is the negative of the sum of zeroes.
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\), इसलिए योग \(3\sqrt{3}\) है। एकक बहुपद में (x) का गुणांक शून्यकों के योग का ऋणात्मक होता है।
A. \(\sqrt{5}\) और \(\sqrt{7}\)/\(\sqrt{5}\) and \(\sqrt{7}\)
Step 1
Concept
The sum is \(\sqrt{5}+\sqrt{7}\) and the product is \(\sqrt{35}\). Both match \(\sqrt{5}\) and \(\sqrt{7}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{5}\) और \(\sqrt{7}\) / \(\sqrt{5}\) and \(\sqrt{7}\). The sum is \(\sqrt{5}+\sqrt{7}\) and the product is \(\sqrt{35}\). Both match \(\sqrt{5}\) and \(\sqrt{7}\).
Step 3
Exam Tip
योग \(\sqrt{5}+\sqrt{7}\) और गुणनफल \(\sqrt{35}\) है। ये दोनों \(\sqrt{5}\) और \(\sqrt{7}\) से मिलते हैं।
A. दोनों शून्यक \(\sqrt{10}\) हैं/Both zeroes are \(\sqrt{10}\)
Step 1
Concept
(p(x)=\(x-\sqrt{10}\)2), so the zero \(\sqrt{10}\) occurs twice. A perfect-square form quickly gives equal zeroes.
Step 2
Why this answer is correct
The correct answer is A. दोनों शून्यक \(\sqrt{10}\) हैं / Both zeroes are \(\sqrt{10}\). (p(x)=\(x-\sqrt{10}\)2), so the zero \(\sqrt{10}\) occurs twice. A perfect-square form quickly gives equal zeroes.
Step 3
Exam Tip
(p(x)=\(x-\sqrt{10}\)2), इसलिए शून्यक दो बार \(\sqrt{10}\) है। पूर्ण वर्ग रूप से समान शून्यक तुरंत मिलते हैं।
\(\alpha+\beta=4\) and \(\alpha\beta=-1\), so \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{4}{-1}=-4\). Find sum and product first.
Step 2
Why this answer is correct
The correct answer is A. (-4). \(\alpha+\beta=4\) and \(\alpha\beta=-1\), so \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{4}{-1}=-4\). Find sum and product first.
Step 3
Exam Tip
\(\alpha+\beta=4\) और \(\alpha\beta=-1\), इसलिए \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{4}{-1}=-4\)। पहले योग और गुणनफल निकालें।
For \(x^2-8x+3\), (D=64-12=52), positive and not a perfect square. The other options give equal rational, non-real, or rational zeroes.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-8x+3\). For \(x^2-8x+3\), (D=64-12=52), positive and not a perfect square. The other options give equal rational, non-real, or rational zeroes.
Step 3
Exam Tip
\(x^2-8x+3\) के लिए (D=64-12=52), जो धनात्मक अपूर्ण वर्ग है। बाकी विकल्पों में शून्यक समान परिमेय, अवास्तविक या परिमेय हैं।
\(\alpha+\beta=2\) and \(\alpha\beta=-1\), so (\alpha-3+\beta-3=23-3(-1)(2)=14). Use (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)).
Step 2
Why this answer is correct
The correct answer is A. (14). \(\alpha+\beta=2\) and \(\alpha\beta=-1\), so (\alpha-3+\beta-3=23-3(-1)(2)=14). Use (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)).
Step 3
Exam Tip
\(\alpha+\beta=2\) और \(\alpha\beta=-1\), इसलिए (\alpha-3+\beta-3=23-3(-1)(2)=14)। घन योग में (\alpha-3+\beta-3=\(\alpha+\beta\)3-3\alpha\beta\(\alpha+\beta\)) लगाएँ।
A. (p(x)) के शून्यक परिमेय हैं और (q(x)) के शून्यक अपरिमेय वास्तविक हैं/(p(x)) has rational zeroes and (q(x)) has irrational real zeroes
Step 1
Concept
For (p(x)), (D=4+32=36), a perfect square. For (q(x)), (D=4+28=32), positive and not a perfect square.
Step 2
Why this answer is correct
The correct answer is A. (p(x)) के शून्यक परिमेय हैं और (q(x)) के शून्यक अपरिमेय वास्तविक हैं / (p(x)) has rational zeroes and (q(x)) has irrational real zeroes. For (p(x)), (D=4+32=36), a perfect square. For (q(x)), (D=4+28=32), positive and not a perfect square.
Step 3
Exam Tip
(p(x)) के लिए (D=4+32=36) पूर्ण वर्ग है। (q(x)) के लिए (D=4+28=32) धनात्मक अपूर्ण वर्ग है।
By the formula, \(x=\frac{8\pm\sqrt{64-8}}{4}=2\pm\frac{\sqrt{14}}{2}\). Divide the whole numerator by the denominator carefully.
Step 2
Why this answer is correct
The correct answer is A. \(2\pm\frac{\sqrt{14}}{2}\). By the formula, \(x=\frac{8\pm\sqrt{64-8}}{4}=2\pm\frac{\sqrt{14}}{2}\). Divide the whole numerator by the denominator carefully.
Step 3
Exam Tip
सूत्र से \(x=\frac{8\pm\sqrt{64-8}}{4}=2\pm\frac{\sqrt{14}}{2}\) है। हर से भाग देते समय पूरे अंश को बाँटें।
\(\alpha+\beta=2\) and \(\alpha\beta=-4\), so (\alpha-2+\beta-2=22-2(-4)=12). Symmetric values can be found without finding the zeroes.
Step 2
Why this answer is correct
The correct answer is A. (12). \(\alpha+\beta=2\) and \(\alpha\beta=-4\), so (\alpha-2+\beta-2=22-2(-4)=12). Symmetric values can be found without finding the zeroes.
Step 3
Exam Tip
\(\alpha+\beta=2\) और \(\alpha\beta=-4\), इसलिए (\alpha-2+\beta-2=22-2(-4)=12)। शून्यक निकाले बिना सममित मान निकाल सकते हैं।
\(\sqrt{8}=2\sqrt{2}\), so the sum is \(\sqrt{2}-2\sqrt{2}=-\sqrt{2}\). Simplifying radicals first reduces mistakes.
Step 2
Why this answer is correct
The correct answer is A. \(-\sqrt{2}\). \(\sqrt{8}=2\sqrt{2}\), so the sum is \(\sqrt{2}-2\sqrt{2}=-\sqrt{2}\). Simplifying radicals first reduces mistakes.
Step 3
Exam Tip
\(\sqrt{8}=2\sqrt{2}\), इसलिए योग \(\sqrt{2}-2\sqrt{2}=-\sqrt{2}\) है। मूलों को पहले सरल करने से गलती कम होती है।
Using the formula, \(x=\frac{2\sqrt{3}\pm\sqrt{12+4}}{2}=\sqrt{3}\pm2\). Simplify \(\sqrt{16}=4\) carefully.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{3}\pm2\). Using the formula, \(x=\frac{2\sqrt{3}\pm\sqrt{12+4}}{2}=\sqrt{3}\pm2\). Simplify \(\sqrt{16}=4\) carefully.
Step 3
Exam Tip
सूत्र से \(x=\frac{2\sqrt{3}\pm\sqrt{12+4}}{2}=\sqrt{3}\pm2\)। \(\sqrt{16}=4\) को ध्यान से सरल करें।
The product is \(ab=\sqrt{2}\cdot\sqrt{18}=\sqrt{36}=6\). In radical multiplication, simplify the product inside the root first.
Step 2
Why this answer is correct
The correct answer is A. (6). The product is \(ab=\sqrt{2}\cdot\sqrt{18}=\sqrt{36}=6\). In radical multiplication, simplify the product inside the root first.
Step 3
Exam Tip
गुणनफल \(ab=\sqrt{2}\cdot\sqrt{18}=\sqrt{36}=6\) है। मूलों के गुणन में पहले अंदर के गुणनफल को सरल करें।
A. \(-2+\sqrt{2}\) और \(-2-\sqrt{2}\)/\(-2+\sqrt{2}\) and \(-2-\sqrt{2}\)
Step 1
Concept
By the formula, \(x=\frac{-4\pm\sqrt{16-8}}{2}=-2\pm\sqrt{2}\). Pay attention to the negative sign and denominator (2).
Step 2
Why this answer is correct
The correct answer is A. \(-2+\sqrt{2}\) और \(-2-\sqrt{2}\) / \(-2+\sqrt{2}\) and \(-2-\sqrt{2}\). By the formula, \(x=\frac{-4\pm\sqrt{16-8}}{2}=-2\pm\sqrt{2}\). Pay attention to the negative sign and denominator (2).
Step 3
Exam Tip
सूत्र से \(x=\frac{-4\pm\sqrt{16-8}}{2}=-2\pm\sqrt{2}\)। ऋण चिह्न और हर (2) दोनों पर ध्यान दें।
(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=32-4(-2)=17). This method gives the answer without finding the zeroes.
Step 2
Why this answer is correct
The correct answer is A. (17). (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=32-4(-2)=17). This method gives the answer without finding the zeroes.
Step 3
Exam Tip
(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=32-4(-2)=17)। यह तरीका शून्यक निकाले बिना उत्तर देता है।
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।
A. \(b^2-4c\) धनात्मक अपूर्ण वर्ग हो/\(b^2-4c\) is positive and not a perfect square
Step 1
Concept
For real zeroes, the discriminant must be positive, and for irrational zeroes it must not be a perfect square. This is the key check for quadratics with rational coefficients.
Step 2
Why this answer is correct
The correct answer is A. \(b^2-4c\) धनात्मक अपूर्ण वर्ग हो / \(b^2-4c\) is positive and not a perfect square. For real zeroes, the discriminant must be positive, and for irrational zeroes it must not be a perfect square. This is the key check for quadratics with rational coefficients.
Step 3
Exam Tip
वास्तविक शून्यकों के लिए विविक्तकर धनात्मक चाहिए और अपरिमेय शून्यकों के लिए वह पूर्ण वर्ग नहीं होना चाहिए। परिमेय गुणांकों वाले द्विघात में यही मुख्य जाँच है।
\(\alpha+\beta=2\) and \(\alpha\beta=-11\), so (\alpha-2+\beta-2+\alpha\beta=\(\alpha+\beta\)2-\alpha\beta=4+11=15). Sum and product are enough for symmetric expressions.
Step 2
Why this answer is correct
The correct answer is A. (15). \(\alpha+\beta=2\) and \(\alpha\beta=-11\), so (\alpha-2+\beta-2+\alpha\beta=\(\alpha+\beta\)2-\alpha\beta=4+11=15). Sum and product are enough for symmetric expressions.
Step 3
Exam Tip
\(\alpha+\beta=2\) और \(\alpha\beta=-11\), इसलिए (\alpha-2+\beta-2+\alpha\beta=\(\alpha+\beta\)2-\alpha\beta=4+11=15)। सममित व्यंजकों में योग और गुणनफल काफी होते हैं।
The sum is \(3+\sqrt{2}\) and the product is \(3\sqrt{2}\). These match (3) and \(\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. (3) और \(\sqrt{2}\) / (3) and \(\sqrt{2}\). The sum is \(3+\sqrt{2}\) and the product is \(3\sqrt{2}\). These match (3) and \(\sqrt{2}\).
Step 3
Exam Tip
योग \(3+\sqrt{2}\) और गुणनफल \(3\sqrt{2}\) है। ये (3) और \(\sqrt{2}\) से मिलते हैं।
A. दोनों परिमेय वास्तविक हैं/Both are rational real
Step 1
Concept
From \(x^2-16=0\), \(x=\pm4\), which are rational real. Not every square-root type question gives irrational roots.
Step 2
Why this answer is correct
The correct answer is A. दोनों परिमेय वास्तविक हैं / Both are rational real. From \(x^2-16=0\), \(x=\pm4\), which are rational real. Not every square-root type question gives irrational roots.
Step 3
Exam Tip
\(x^2-16=0\) से \(x=\pm4\), जो परिमेय वास्तविक हैं। हर वर्गमूल वाला प्रश्न अपरिमेय नहीं होता।
The zeroes are \(3\sqrt{2}\) and \(-3\sqrt{2}\). Therefore the sum is (0) and the product is (-18).
Step 2
Why this answer is correct
The correct answer is A. गुणनफल (-18), योग (0) / Product (-18), sum (0). The zeroes are \(3\sqrt{2}\) and \(-3\sqrt{2}\). Therefore the sum is (0) and the product is (-18).
Step 3
Exam Tip
शून्यक \(3\sqrt{2}\) और \(-3\sqrt{2}\) हैं। इसलिए योग (0) और गुणनफल (-18) है।
A. \(-\sqrt{7}+1\) और \(-\sqrt{7}-1\)/\(-\sqrt{7}+1\) and \(-\sqrt{7}-1\)
Step 1
Concept
Using the formula, \(x=\frac{-2\sqrt{7}\pm2}{2}=-\sqrt{7}\pm1\). Simplifying the discriminant first gives a clean answer.
Step 2
Why this answer is correct
The correct answer is A. \(-\sqrt{7}+1\) और \(-\sqrt{7}-1\) / \(-\sqrt{7}+1\) and \(-\sqrt{7}-1\). Using the formula, \(x=\frac{-2\sqrt{7}\pm2}{2}=-\sqrt{7}\pm1\). Simplifying the discriminant first gives a clean answer.
Step 3
Exam Tip
सूत्र से \(x=\frac{-2\sqrt{7}\pm2}{2}=-\sqrt{7}\pm1\)। पहले विविक्तकर सरल करने से उत्तर साफ मिलता है।
A. ऐसा कोई वास्तविक (n) नहीं है/No such real (n) exists
Step 1
Concept
For equal zeroes, (D=0), so (4-4n=0) and (n=1). Then the zero is (1), which is not irrational.
Step 2
Why this answer is correct
The correct answer is A. ऐसा कोई वास्तविक (n) नहीं है / No such real (n) exists. For equal zeroes, (D=0), so (4-4n=0) and (n=1). Then the zero is (1), which is not irrational.
Step 3
Exam Tip
समान शून्यकों के लिए (D=0), यानी (4-4n=0), इसलिए (n=1)। तब शून्यक (1) है, जो अपरिमेय नहीं है।
With rational coefficients, the conjugate of the irrational part is also a zero. Hence \(\frac{3-\sqrt{5}}{2}\) is the other zero.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3-\sqrt{5}}{2}\). With rational coefficients, the conjugate of the irrational part is also a zero. Hence \(\frac{3-\sqrt{5}}{2}\) is the other zero.
Step 3
Exam Tip
परिमेय गुणांकों में अपरिमेय भाग का संयुग्मी भी शून्यक होता है। इसलिए \(\frac{3-\sqrt{5}}{2}\) दूसरा शून्यक है।
A. दो भिन्न वास्तविक अपरिमेय/Two distinct real irrational
Step 1
Concept
(D=\(2\sqrt{2}\)2-4=8-4=4), and the zeroes are \(\sqrt{2}\pm1\). They are real and irrational.
Step 2
Why this answer is correct
The correct answer is A. दो भिन्न वास्तविक अपरिमेय / Two distinct real irrational. (D=\(2\sqrt{2}\)2-4=8-4=4), and the zeroes are \(\sqrt{2}\pm1\). They are real and irrational.
Step 3
Exam Tip
(D=\(2\sqrt{2}\)2-4=8-4=4) है और शून्यक \(\sqrt{2}\pm1\) हैं। ये वास्तविक और अपरिमेय हैं।
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. (p(x)) के शून्यक परिमेय और (q(x)) के अपरिमेय वास्तविक हैं/(p(x)) has rational zeroes and (q(x)) has irrational real zeroes
Step 1
Concept
For (p(x)), (D=121-96=25), a perfect square. For (q(x)), (D=121-92=29), positive and not a perfect square.
Step 2
Why this answer is correct
The correct answer is A. (p(x)) के शून्यक परिमेय और (q(x)) के अपरिमेय वास्तविक हैं / (p(x)) has rational zeroes and (q(x)) has irrational real zeroes. For (p(x)), (D=121-96=25), a perfect square. For (q(x)), (D=121-92=29), positive and not a perfect square.
Step 3
Exam Tip
(p(x)) के लिए (D=121-96=25) पूर्ण वर्ग है। (q(x)) के लिए (D=121-92=29) धनात्मक अपूर्ण वर्ग 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।
A. योग \(6\sqrt{2}\), गुणनफल (17)/Sum \(6\sqrt{2}\), product (17)
Step 1
Concept
In a monic quadratic, the sum is (-b) and the product is (c). Therefore the sum is \(6\sqrt{2}\) and the product is (17).
Step 2
Why this answer is correct
The correct answer is A. योग \(6\sqrt{2}\), गुणनफल (17) / Sum \(6\sqrt{2}\), product (17). In a monic quadratic, the sum is (-b) and the product is (c). Therefore the sum is \(6\sqrt{2}\) and the product is (17).
Step 3
Exam Tip
एकक द्विघात में योग (-b) और गुणनफल (c) होता है। इसलिए योग \(6\sqrt{2}\) और गुणनफल (17) है।
The sum is \(4\sqrt{3}\) and the product is (\(2\sqrt{3}\)2-1=11). (S) equals the sum and (P) equals the product.
Step 2
Why this answer is correct
The correct answer is A. \(S=4\sqrt{3}\), (P=11). The sum is \(4\sqrt{3}\) and the product is (\(2\sqrt{3}\)2-1=11). (S) equals the sum and (P) equals the product.
Step 3
Exam Tip
योग \(4\sqrt{3}\) और गुणनफल (\(2\sqrt{3}\)2-1=11) है। (S) योग और (P) गुणनफल के बराबर है।
A. शून्यक \(6+\sqrt{5}\) और \(6-\sqrt{5}\)/Zeroes \(6+\sqrt{5}\) and \(6-\sqrt{5}\)
Step 1
Concept
With rational coefficients, irrational parts occur in conjugate pairs. Only \(6+\sqrt{5}\) and \(6-\sqrt{5}\) have both rational sum and rational product.
Step 2
Why this answer is correct
The correct answer is A. शून्यक \(6+\sqrt{5}\) और \(6-\sqrt{5}\) / Zeroes \(6+\sqrt{5}\) and \(6-\sqrt{5}\). With rational coefficients, irrational parts occur in conjugate pairs. Only \(6+\sqrt{5}\) and \(6-\sqrt{5}\) have both rational sum and rational product.
Step 3
Exam Tip
परिमेय गुणांकों में अपरिमेय भाग संयुग्मी जोड़े में आता है। केवल \(6+\sqrt{5}\) और \(6-\sqrt{5}\) का योग और गुणनफल दोनों परिमेय हैं।
(\(1+\sqrt{3}\)2-2\(1+\sqrt{3}\)-2=1+2\sqrt{3}+3-2-2\sqrt{3}-2=0). Do not forget the middle term while expanding the square.
Step 2
Why this answer is correct
The correct answer is A. (0). (\(1+\sqrt{3}\)2-2\(1+\sqrt{3}\)-2=1+2\sqrt{3}+3-2-2\sqrt{3}-2=0). Do not forget the middle term while expanding the square.
Step 3
Exam Tip
(\(1+\sqrt{3}\)2-2\(1+\sqrt{3}\)-2=1+2\sqrt{3}+3-2-2\sqrt{3}-2=0)। वर्ग खोलते समय बीच का पद न भूलें।
Since (p\(1+\sqrt{3}\)=0), \(1+\sqrt{3}\) is a zero. To prove a number is a zero, show that the polynomial value is (0).
Step 2
Why this answer is correct
The correct answer is A. यह (p(x)) का शून्यक है / It is a zero of (p(x)). Since (p\(1+\sqrt{3}\)=0), \(1+\sqrt{3}\) is a zero. To prove a number is a zero, show that the polynomial value is (0).
Step 3
Exam Tip
(p\(1+\sqrt{3}\)=0), इसलिए \(1+\sqrt{3}\) शून्यक है। किसी संख्या को शून्यक सिद्ध करने के लिए बहुपद का मान (0) दिखाएँ।
The sum of zeroes is (2), so the other zero is (2-\(1+\sqrt{3}\)=1-\sqrt{3}). With rational coefficients, the conjugate also appears.
Step 2
Why this answer is correct
The correct answer is A. \(1-\sqrt{3}\). The sum of zeroes is (2), so the other zero is (2-\(1+\sqrt{3}\)=1-\sqrt{3}). With rational coefficients, the conjugate also appears.
Step 3
Exam Tip
शून्यकों का योग (2) है, इसलिए दूसरा शून्यक (2-\(1+\sqrt{3}\)=1-\sqrt{3}) है। परिमेय गुणांकों में संयुग्मी भी मिलता है।
Here \(\alpha+\beta=8\) and \(\alpha\beta=1\), so \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{64-2}{1}=62\). In such questions, first find the sum and product.
Step 2
Why this answer is correct
The correct answer is A. (62). Here \(\alpha+\beta=8\) and \(\alpha\beta=1\), so \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{64-2}{1}=62\). In such questions, first find the sum and product.
Step 3
Exam Tip
\(\alpha+\beta=8\) और \(\alpha\beta=1\), इसलिए \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}=\frac{64-2}{1}=62\)। ऐसे प्रश्नों में पहले योग और गुणनफल निकालें।
