The other zero will be \(6-\sqrt{19}\), and the product is (36-19=17). In exams connect the constant term with the product of zeroes.
Step 2
Why this answer is correct
The correct answer is A. (17). The other zero will be \(6-\sqrt{19}\), and the product is (36-19=17). In exams connect the constant term with the product of zeroes.
Step 3
Exam Tip
दूसरा शून्यक \(6-\sqrt{19}\) होगा और गुणनफल (36-19=17) है। परीक्षा में स्थिर पद को शून्यकों के गुणनफल से जोड़ें।
The discriminant is (196-152=44) and \(\sqrt{44}\) is irrational. Hence the zeroes are real irrational.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक अपरिमेय / Real irrational. The discriminant is (196-152=44) and \(\sqrt{44}\) is irrational. Hence the zeroes are real irrational.
Step 3
Exam Tip
विविक्तकर (196-152=44) है और \(\sqrt{44}\) अपरिमेय है। इसलिए शून्यक वास्तविक अपरिमेय हैं।
The sum is (12) and the product is (36-11=25), so the polynomial is \(x^2-12x+25\). In exams write the standard form correctly.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-12x+25\). The sum is (12) and the product is (36-11=25), so the polynomial is \(x^2-12x+25\). In exams write the standard form correctly.
Step 3
Exam Tip
योग (12) और गुणनफल (36-11=25) है, इसलिए बहुपद \(x^2-12x+25\) है। परीक्षा में मानक रूप ठीक से लिखें।
The discriminant is (100-92=8), and \(\sqrt{8}\) is irrational, so the zeroes are real irrational. In exams check the square root of the discriminant.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक अपरिमेय / Real irrational. The discriminant is (100-92=8), and \(\sqrt{8}\) is irrational, so the zeroes are real irrational. In exams check the square root of the discriminant.
Step 3
Exam Tip
विविक्तकर (100-92=8) है और \(\sqrt{8}\) अपरिमेय है, इसलिए शून्यक वास्तविक अपरिमेय हैं। परीक्षा में विविक्तकर का वर्गमूल देखें।
With rational coefficients, \(a+\sqrt{b}\) is accompanied by \(a-\sqrt{b}\). In exams identify conjugate zeroes quickly.
Step 2
Why this answer is correct
The correct answer is A. \(2-\sqrt{7}\). With rational coefficients, \(a+\sqrt{b}\) is accompanied by \(a-\sqrt{b}\). In exams identify conjugate zeroes quickly.
Step 3
Exam Tip
परिमेय गुणांकों में \(a+\sqrt{b}\) के साथ \(a-\sqrt{b}\) भी शून्यक आता है। परीक्षा में संयुग्मी शून्यकों को तुरंत पहचानें।
For rational coefficients, irrational zeroes usually occur in conjugate pairs. Hence the companion zero of \(3-\sqrt{5}\) is \(3+\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(3+\sqrt{5}\). For rational coefficients, irrational zeroes usually occur in conjugate pairs. Hence the companion zero of \(3-\sqrt{5}\) is \(3+\sqrt{5}\).
Step 3
Exam Tip
परिमेय गुणांकों में अपरिमेय शून्यक सामान्यतः संयुग्मी रूप में आते हैं। इसलिए \(3-\sqrt{5}\) का साथी शून्यक \(3+\sqrt{5}\) होगा।
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।
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।
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।
A. शून्यक \(a+\sqrt{b}\) और \(a-\sqrt{b}\) हैं, दोनों वास्तविक अपरिमेय हो सकते हैं/Zeroes are \(a+\sqrt{b}\) and \(a-\sqrt{b}\), both can be real irrational
Step 1
Concept
The polynomial equals ((x-a)2-b), so \(x=a\pm\sqrt{b}\). When (b) is not a perfect square, \(\sqrt{b}\) is irrational.
Step 2
Why this answer is correct
The correct answer is A. शून्यक \(a+\sqrt{b}\) और \(a-\sqrt{b}\) हैं, दोनों वास्तविक अपरिमेय हो सकते हैं / Zeroes are \(a+\sqrt{b}\) and \(a-\sqrt{b}\), both can be real irrational. The polynomial equals ((x-a)2-b), so \(x=a\pm\sqrt{b}\). When (b) is not a perfect square, \(\sqrt{b}\) is irrational.
Step 3
Exam Tip
बहुपद ((x-a)2-b) के बराबर है, इसलिए \(x=a\pm\sqrt{b}\) है। जब (b) पूर्ण वर्ग नहीं है तो \(\sqrt{b}\) अपरिमेय होता है।
In \(x^2-4x+1\), the sum is (4) and (D=16-4=12), so the zeroes are irrational. A rational sum does not mean rational zeroes.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4x+1\). In \(x^2-4x+1\), the sum is (4) and (D=16-4=12), so the zeroes are irrational. A rational sum does not mean rational zeroes.
Step 3
Exam Tip
\(x^2-4x+1\) में योग (4) है और (D=16-4=12) से शून्यक अपरिमेय हैं। परिमेय योग का अर्थ परिमेय शून्यक होना नहीं है।
A. \(4+\sqrt{6}\) और \(4-\sqrt{6}\)/\(4+\sqrt{6}\) and \(4-\sqrt{6}\)
Step 1
Concept
For rational coefficients, the conjugate \(a-\sqrt{b}\) accompanies \(a+\sqrt{b}\). Hence the first pair is correct.
Step 2
Why this answer is correct
The correct answer is A. \(4+\sqrt{6}\) और \(4-\sqrt{6}\) / \(4+\sqrt{6}\) and \(4-\sqrt{6}\). For rational coefficients, the conjugate \(a-\sqrt{b}\) accompanies \(a+\sqrt{b}\). Hence the first pair is correct.
Step 3
Exam Tip
परिमेय गुणांकों के लिए \(a+\sqrt{b}\) का संयुग्मी \(a-\sqrt{b}\) साथ आता है। इसलिए पहला युग्म सही है।
A. दो भिन्न अपरिमेय वास्तविक शून्यक/Two distinct irrational real zeroes
Step 1
Concept
The discriminant is (D=36-16=20), so the zeroes are \(3\pm\sqrt{5}\). If (D) is not a perfect square, real zeroes can be irrational.
Step 2
Why this answer is correct
The correct answer is A. दो भिन्न अपरिमेय वास्तविक शून्यक / Two distinct irrational real zeroes. The discriminant is (D=36-16=20), so the zeroes are \(3\pm\sqrt{5}\). If (D) is not a perfect square, real zeroes can be irrational.
Step 3
Exam Tip
विविक्तकर (D=36-16=20) है, इसलिए शून्यक \(3\pm\sqrt{5}\) हैं। (D) पूर्ण वर्ग न हो तो वास्तविक शून्यक अपरिमेय हो सकते हैं।
For (r=2), (D=16-8=8). It is positive and not a perfect square, so the zeroes are real and irrational.
Step 2
Why this answer is correct
The correct answer is A. कथन सही है / The statement is true. For (r=2), (D=16-8=8). It is positive and not a perfect square, so the zeroes are real and irrational.
Step 3
Exam Tip
(r=2) पर (D=16-8=8) है। यह धनात्मक और अपूर्ण वर्ग है, इसलिए शून्यक वास्तविक और अपरिमेय हैं।
A. \(\sqrt{2}\) और \(\sqrt{3}\)/\(\sqrt{2}\) and \(\sqrt{3}\)
Step 1
Concept
The sum \(\sqrt{2}+\sqrt{3}\) and product \(\sqrt{6}\) match the option \(\sqrt{2}\), \(\sqrt{3}\). Hence those are the zeroes.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\) और \(\sqrt{3}\) / \(\sqrt{2}\) and \(\sqrt{3}\). The sum \(\sqrt{2}+\sqrt{3}\) and product \(\sqrt{6}\) match the option \(\sqrt{2}\), \(\sqrt{3}\). Hence those are the zeroes.
Step 3
Exam Tip
योग \(\sqrt{2}+\sqrt{3}\) और गुणनफल \(\sqrt{6}\) विकल्प \(\sqrt{2}\), \(\sqrt{3}\) से मिलते हैं। इसलिए वही शून्यक हैं।
A. दो भिन्न अपरिमेय वास्तविक शून्यक/Two distinct irrational real zeroes
Step 1
Concept
(D=36-28=8). It is positive but not a perfect square, so the zeroes are real and irrational.
Step 2
Why this answer is correct
The correct answer is A. दो भिन्न अपरिमेय वास्तविक शून्यक / Two distinct irrational real zeroes. (D=36-28=8). It is positive but not a perfect square, so the zeroes are real and irrational.
Step 3
Exam Tip
(D=36-28=8) है। (8) धनात्मक है पर पूर्ण वर्ग नहीं, इसलिए शून्यक वास्तविक और अपरिमेय हैं।
B. जब (25-4c) धनात्मक हो पर पूर्ण वर्ग न हो/When (25-4c) is positive but not a perfect square
Step 1
Concept
For real distinct zeroes, (D>0) is required. For irrational zeroes, (D) must not be a perfect square.
Step 2
Why this answer is correct
The correct answer is B. जब (25-4c) धनात्मक हो पर पूर्ण वर्ग न हो / When (25-4c) is positive but not a perfect square. For real distinct zeroes, (D>0) is required. For irrational zeroes, (D) must not be a perfect square.
Step 3
Exam Tip
वास्तविक भिन्न शून्यकों के लिए (D>0) चाहिए। अपरिमेय शून्यकों के लिए (D) पूर्ण वर्ग नहीं होना चाहिए।
A. \(-1+\sqrt{2}\) और \(-1-\sqrt{2}\)/\(-1+\sqrt{2}\) and \(-1-\sqrt{2}\)
Step 1
Concept
By the formula, \(x=\frac{-2\pm\sqrt{4+4}}{2}=-1\pm\sqrt{2}\). Pay special attention to signs.
Step 2
Why this answer is correct
The correct answer is A. \(-1+\sqrt{2}\) और \(-1-\sqrt{2}\) / \(-1+\sqrt{2}\) and \(-1-\sqrt{2}\). By the formula, \(x=\frac{-2\pm\sqrt{4+4}}{2}=-1\pm\sqrt{2}\). Pay special attention to signs.
Step 3
Exam Tip
सूत्र से \(x=\frac{-2\pm\sqrt{4+4}}{2}=-1\pm\sqrt{2}\)। चिह्नों पर विशेष ध्यान दें।
A. \(4+\sqrt{6}\) और \(4-\sqrt{6}\)/\(4+\sqrt{6}\) and \(4-\sqrt{6}\)
Step 1
Concept
By the formula, the zeroes are \(\frac{8\pm\sqrt{64-40}}{2}=4\pm\sqrt{6}\). Simplify the discriminant first.
Step 2
Why this answer is correct
The correct answer is A. \(4+\sqrt{6}\) और \(4-\sqrt{6}\) / \(4+\sqrt{6}\) and \(4-\sqrt{6}\). By the formula, the zeroes are \(\frac{8\pm\sqrt{64-40}}{2}=4\pm\sqrt{6}\). Simplify the discriminant first.
Step 3
Exam Tip
सूत्र से शून्यक \(\frac{8\pm\sqrt{64-40}}{2}=4\pm\sqrt{6}\) हैं। पहले विविक्तकर सरल करें।
A. दोनों शून्यक \(\sqrt{5}\) हैं/Both zeroes are \(\sqrt{5}\)
Step 1
Concept
This polynomial equals (\(x-\sqrt{5}\)2). Hence \(\sqrt{5}\) is a repeated zero.
Step 2
Why this answer is correct
The correct answer is A. दोनों शून्यक \(\sqrt{5}\) हैं / Both zeroes are \(\sqrt{5}\). This polynomial equals (\(x-\sqrt{5}\)2). Hence \(\sqrt{5}\) is a repeated zero.
Step 3
Exam Tip
यह बहुपद (\(x-\sqrt{5}\)2) के बराबर है। अतः \(\sqrt{5}\) दोहरा शून्यक है।
C. दो भिन्न अपरिमेय शून्यक/Two distinct irrational zeroes
Step 1
Concept
The discriminant is (D=36-16=20), and (20) is not a perfect square. So the zeroes are real, distinct, and irrational.
Step 2
Why this answer is correct
The correct answer is C. दो भिन्न अपरिमेय शून्यक / Two distinct irrational zeroes. The discriminant is (D=36-16=20), and (20) is not a perfect square. So the zeroes are real, distinct, and irrational.
Step 3
Exam Tip
विविक्तकर (D=36-16=20) है और (20) पूर्ण वर्ग नहीं है। इसलिए शून्यक वास्तविक भिन्न और अपरिमेय हैं।
