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।
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।
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।
For a downward-opening parabola, values outside the zeroes are negative. Tip: opening direction changes sign regions.
Step 2
Why this answer is correct
The correct answer is B. (x)-अक्ष के नीचे / Below the (x)-axis. For a downward-opening parabola, values outside the zeroes are negative. Tip: opening direction changes sign regions.
Step 3
Exam Tip
नीचे खुलने वाले परवलय में शून्यकों के बाहर मान ऋणात्मक होते हैं। टिप: खुलने की दिशा संकेत क्षेत्र बदलती है।
For a downward-opening parabola, values outside the zeroes are negative. Tip: opening direction changes sign regions.
Step 2
Why this answer is correct
The correct answer is B. (x)-अक्ष के नीचे / Below the (x)-axis. For a downward-opening parabola, values outside the zeroes are negative. Tip: opening direction changes sign regions.
Step 3
Exam Tip
नीचे खुलने वाले परवलय में शून्यकों के बाहर मान ऋणात्मक होते हैं। टिप: खुलने की दिशा संकेत क्षेत्र बदलती है।
For a downward-opening parabola, values outside the zeroes are negative. Tip: when the direction changes, sign regions also change.
Step 2
Why this answer is correct
The correct answer is B. (x)-अक्ष के नीचे / Below the (x)-axis. For a downward-opening parabola, values outside the zeroes are negative. Tip: when the direction changes, sign regions also change.
Step 3
Exam Tip
नीचे खुलने वाले परवलय में शून्यकों के बाहर मान ऋणात्मक होते हैं। टिप: दिशा बदलने पर संकेत क्षेत्र भी बदलता है।
The repeated (2) is counted once for distinct zeroes. Tip: do not rewrite the same value for distinct zeroes.
Step 2
Why this answer is correct
The correct answer is A. (2) और (-5) / (2) and (-5). The repeated (2) is counted once for distinct zeroes. Tip: do not rewrite the same value for distinct zeroes.
Step 3
Exam Tip
दोहराया (2) अलग शून्यक में एक बार गिना जाता है। टिप: अलग शून्यक में समान मान पुनः न लिखें।
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।
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।
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।
B. (k) धनात्मक हो लेकिन पूर्ण वर्ग न हो/(k) is positive but not a perfect square
Step 1
Concept
The zeroes are \(x=\pm\sqrt{k}\). They are irrational real when (k>0) and (k) is not a perfect square.
Step 2
Why this answer is correct
The correct answer is B. (k) धनात्मक हो लेकिन पूर्ण वर्ग न हो / (k) is positive but not a perfect square. The zeroes are \(x=\pm\sqrt{k}\). They are irrational real when (k>0) and (k) is not a perfect square.
Step 3
Exam Tip
शून्यक \(x=\pm\sqrt{k}\) हैं। ये अपरिमेय वास्तविक तभी होंगे जब (k>0) और (k) पूर्ण वर्ग न हो।
A. \(4+\sqrt{5}\) और \(4-\sqrt{5}\)/\(4+\sqrt{5}\) and \(4-\sqrt{5}\)
Step 1
Concept
The sum of \(4+\sqrt{5}\) and \(4-\sqrt{5}\) is (8), and the product is (16-5=11). In exams check the sum and product of options.
Step 2
Why this answer is correct
The correct answer is A. \(4+\sqrt{5}\) और \(4-\sqrt{5}\) / \(4+\sqrt{5}\) and \(4-\sqrt{5}\). The sum of \(4+\sqrt{5}\) and \(4-\sqrt{5}\) is (8), and the product is (16-5=11). In exams check the sum and product of options.
Step 3
Exam Tip
\(4+\sqrt{5}\) और \(4-\sqrt{5}\) का योग (8) और गुणनफल (16-5=11) है। परीक्षा में विकल्पों का योग और गुणनफल जांचें।
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।
A. \(4+\sqrt{3}\) और \(4-\sqrt{3}\)/\(4+\sqrt{3}\) and \(4-\sqrt{3}\)
Step 1
Concept
Using the quadratic formula \(x=\frac{8\pm\sqrt{64-52}}{2}=4\pm\sqrt{3}\). In exams simplify the discriminant.
Step 2
Why this answer is correct
The correct answer is A. \(4+\sqrt{3}\) और \(4-\sqrt{3}\) / \(4+\sqrt{3}\) and \(4-\sqrt{3}\). Using the quadratic formula \(x=\frac{8\pm\sqrt{64-52}}{2}=4\pm\sqrt{3}\). In exams simplify the discriminant.
Step 3
Exam Tip
द्विघात सूत्र से \(x=\frac{8\pm\sqrt{64-52}}{2}=4\pm\sqrt{3}\) मिलता है। परीक्षा में विविक्तकर को सरल करें।
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।
A. \(2+\sqrt{5}\) और \(2-\sqrt{5}\)/\(2+\sqrt{5}\) and \(2-\sqrt{5}\)
Step 1
Concept
Using the quadratic formula, \(x=\frac{4\pm\sqrt{16+4}}{2}=2\pm\sqrt{5}\). In exams simplify the discriminant.
Step 2
Why this answer is correct
The correct answer is A. \(2+\sqrt{5}\) और \(2-\sqrt{5}\) / \(2+\sqrt{5}\) and \(2-\sqrt{5}\). Using the quadratic formula, \(x=\frac{4\pm\sqrt{16+4}}{2}=2\pm\sqrt{5}\). In exams simplify the discriminant.
Step 3
Exam Tip
द्विघात सूत्र से \(x=\frac{4\pm\sqrt{16+4}}{2}=2\pm\sqrt{5}\) मिलता है। परीक्षा में विविक्तकर को सरल करें।
(p(x)=x\(x-\sqrt{5}\)), so the zeroes are (0) and \(\sqrt{5}\). Taking the common factor is a fast method in exams.
Step 2
Why this answer is correct
The correct answer is A. \(0,\sqrt{5}\). (p(x)=x\(x-\sqrt{5}\)), so the zeroes are (0) and \(\sqrt{5}\). Taking the common factor is a fast method in exams.
Step 3
Exam Tip
(p(x)=x\(x-\sqrt{5}\)), इसलिए शून्यक (0) और \(\sqrt{5}\) हैं। परीक्षा में सामान्य गुणनखंड निकालना तेज तरीका है।
The sum is \(1+\sqrt{3}\) and the product is \(\sqrt{3}\), so the zeroes are (1) and \(\sqrt{3}\). Compare with \(x^2-Sx+P\) in exams.
Step 2
Why this answer is correct
The correct answer is A. \(1,\sqrt{3}\). The sum is \(1+\sqrt{3}\) and the product is \(\sqrt{3}\), so the zeroes are (1) and \(\sqrt{3}\). Compare with \(x^2-Sx+P\) in exams.
Step 3
Exam Tip
योग \(1+\sqrt{3}\) और गुणनफल \(\sqrt{3}\) है, इसलिए शून्यक (1) और \(\sqrt{3}\) हैं। परीक्षा में \(x^2-Sx+P\) से तुलना करें।
The discriminant is (100-76=24), so the zeroes are \(\frac{10\pm\sqrt{24}}{2}=5\pm\sqrt{6}\). Simplify \(\sqrt{24}=2\sqrt{6}\) in exams.
Step 2
Why this answer is correct
The correct answer is A. \(5+\sqrt{6},5-\sqrt{6}\). The discriminant is (100-76=24), so the zeroes are \(\frac{10\pm\sqrt{24}}{2}=5\pm\sqrt{6}\). Simplify \(\sqrt{24}=2\sqrt{6}\) in exams.
Step 3
Exam Tip
विविक्तकर (100-76=24) है, इसलिए शून्यक \(\frac{10\pm\sqrt{24}}{2}=5\pm\sqrt{6}\) हैं। परीक्षा में \(\sqrt{24}=2\sqrt{6}\) सरल करें।
The sum is (4) and the product is (4-6=-2), so the polynomial is \(x^2-4x-2\). Remember the formula \(x^2-Sx+P\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4x-2\). The sum is (4) and the product is (4-6=-2), so the polynomial is \(x^2-4x-2\). Remember the formula \(x^2-Sx+P\).
Step 3
Exam Tip
योग (4) और गुणनफल (4-6=-2) है, इसलिए बहुपद \(x^2-4x-2\) है। परीक्षा में \(x^2-Sx+P\) सूत्र याद रखें।
A. शून्यकों का गुणनफल \(-3\sqrt{2}\) है/The product of zeroes is \(-3\sqrt{2}\)
Step 1
Concept
In a monic quadratic, the constant term is the product of zeroes. Here \(\alpha\beta=-3\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. शून्यकों का गुणनफल \(-3\sqrt{2}\) है / The product of zeroes is \(-3\sqrt{2}\). In a monic quadratic, the constant term is the product of zeroes. Here \(\alpha\beta=-3\sqrt{2}\).
Step 3
Exam Tip
एकक द्विघात में स्थिर पद शून्यकों का गुणनफल होता है। यहाँ \(\alpha\beta=-3\sqrt{2}\) है।
The sum is \(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\). Simplify radicals before giving the final answer.
Step 2
Why this answer is correct
The correct answer is A. \(3\sqrt{2}\). The sum is \(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\). Simplify radicals before giving the final answer.
Step 3
Exam Tip
योग \(\sqrt{2}+\sqrt{8}=\sqrt{2}+2\sqrt{2}=3\sqrt{2}\) है। मूलों को सरल करके ही अंतिम उत्तर दें।
A. \(-\sqrt{7}+1\) 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।
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।
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।
\(\sqrt{8}=2\sqrt{2}\), 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।
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।
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।
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।
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।
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. (p(x)) के शून्यक परिमेय हैं और (q(x)) के शून्यक अपरिमेय वास्तविक हैं/Zeroes of (p(x)) are rational and zeroes of (q(x)) are irrational real
Step 1
Concept
For (p(x)), (D=81-56=25), a perfect square, so the zeroes are rational. For (q(x)), (D=81-60=21), positive but not a perfect square, so the zeroes are irrational real.
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 / Zeroes of (p(x)) are rational and zeroes of (q(x)) are irrational real. For (p(x)), (D=81-56=25), a perfect square, so the zeroes are rational. For (q(x)), (D=81-60=21), positive but not a perfect square, so the zeroes are irrational real.
Step 3
Exam Tip
(p(x)) के लिए (D=81-56=25) पूर्ण वर्ग है, इसलिए शून्यक परिमेय हैं। (q(x)) के लिए (D=81-60=21) धनात्मक अपूर्ण वर्ग है, इसलिए शून्यक अपरिमेय वास्तविक हैं।
A. \(\sqrt{2}\) और \(\sqrt{3}\)/\(\sqrt{2}\) and \(\sqrt{3}\)
Step 1
Concept
The sum is \(\sqrt{2}+\sqrt{3}\) and the product is \(\sqrt{6}\). These match \(\sqrt{2}\) and \(\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\) और \(\sqrt{3}\) / \(\sqrt{2}\) and \(\sqrt{3}\). The sum is \(\sqrt{2}+\sqrt{3}\) and the product is \(\sqrt{6}\). These match \(\sqrt{2}\) and \(\sqrt{3}\).
Step 3
Exam Tip
योग \(\sqrt{2}+\sqrt{3}\) और गुणनफल \(\sqrt{6}\) है। ये \(\sqrt{2}\) और \(\sqrt{3}\) से मिलते हैं।
The zeroes are \(6\pm\sqrt{5}\), so the difference is \(2\sqrt{5}\). The difference of conjugate zeroes is (2) times the radical part.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{5}\). The zeroes are \(6\pm\sqrt{5}\), so the difference is \(2\sqrt{5}\). The difference of conjugate zeroes is (2) times the radical part.
Step 3
Exam Tip
शून्यक \(6\pm\sqrt{5}\) हैं, इसलिए अंतर \(2\sqrt{5}\) है। संयुग्मी शून्यकों का अंतर (2) गुणा मूल पद होता है।
The sum of zeroes is \(1+\sqrt{3}\) and the product is \(\sqrt{3}\). The numbers (1) and \(\sqrt{3}\) satisfy both conditions.
Step 2
Why this answer is correct
The correct answer is A. (1) और \(\sqrt{3}\) / (1) and \(\sqrt{3}\). The sum of zeroes is \(1+\sqrt{3}\) and the product is \(\sqrt{3}\). The numbers (1) and \(\sqrt{3}\) satisfy both conditions.
Step 3
Exam Tip
शून्यकों का योग \(1+\sqrt{3}\) और गुणनफल \(\sqrt{3}\) है। (1) और \(\sqrt{3}\) दोनों शर्तें पूरी करते हैं।
The product is (\frac{\(1+\sqrt{3}\)\(1-\sqrt{3}\)}{4}=\frac{1-3}{4}=-\frac{1}{2}). Use \(a^2-b\) for conjugate products.
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{1}{2}\). The product is (\frac{\(1+\sqrt{3}\)\(1-\sqrt{3}\)}{4}=\frac{1-3}{4}=-\frac{1}{2}). Use \(a^2-b\) for conjugate products.
Step 3
Exam Tip
गुणनफल (\frac{\(1+\sqrt{3}\)\(1-\sqrt{3}\)}{4}=\frac{1-3}{4}=-\frac{1}{2}) है। संयुग्मी गुणनफल में \(a^2-b\) प्रयोग करें।
A. योग (-4), दोनों अपरिमेय वास्तविक/Sum (-4), both irrational real
Step 1
Concept
The sum is \(-\frac{b}{a}=-4\) and (D=16-4=12), not a perfect square. Hence both zeroes are irrational real.
Step 2
Why this answer is correct
The correct answer is A. योग (-4), दोनों अपरिमेय वास्तविक / Sum (-4), both irrational real. The sum is \(-\frac{b}{a}=-4\) and (D=16-4=12), not a perfect square. Hence both zeroes are irrational real.
Step 3
Exam Tip
योग \(-\frac{b}{a}=-4\) और (D=16-4=12) है, जो पूर्ण वर्ग नहीं है। इसलिए दोनों शून्यक अपरिमेय वास्तविक हैं।
Since (3x-2-12x+6=3\(x^2-4x+2\)), the zeroes are \(2\pm\sqrt{2}\). Removing a common factor first makes calculation easier.
Step 2
Why this answer is correct
The correct answer is A. \(2\pm\sqrt{2}\). Since (3x-2-12x+6=3\(x^2-4x+2\)), the zeroes are \(2\pm\sqrt{2}\). Removing a common factor first makes calculation easier.
Step 3
Exam Tip
(3x-2-12x+6=3\(x^2-4x+2\)), इसलिए शून्यक \(2\pm\sqrt{2}\) हैं। पहले सामान्य गुणनखंड हटाना गणना आसान करता है।
B. दो समान अपरिमेय शून्यक हैं/It has two equal irrational zeroes
Step 1
Concept
Since (p(x)=\(x-\sqrt{3}\)2), both zeroes are \(\sqrt{3}\). Recognize perfect-square form for equal zeroes.
Step 2
Why this answer is correct
The correct answer is B. दो समान अपरिमेय शून्यक हैं / It has two equal irrational zeroes. Since (p(x)=\(x-\sqrt{3}\)2), both zeroes are \(\sqrt{3}\). Recognize perfect-square form for equal zeroes.
Step 3
Exam Tip
(p(x)=\(x-\sqrt{3}\)2), इसलिए दोनों शून्यक \(\sqrt{3}\) हैं। समान शून्यक के लिए पूर्ण वर्ग रूप पहचानें।
\(The sum of zeroes is (0) and product is (-7), so the polynomial is (x^2-7). Use (x^2-\)sumx+product) to form a polynomial from zeroes.
Step 2
Why this answer is correct
\(The correct answer is B. (x^2-7). The sum of zeroes is (0) and product is (-7), so the polynomial is (x^2-7). Use (x^2-\)sumx+product) to form a polynomial from zeroes.
Step 3
Exam Tip
शून्यकों का योग (0) और गुणनफल (-7) है, इसलिए बहुपद \(x^2-7\) है। \(शून्यकों से बहुपद बनाते समय (x^2-\)योगx+गुणनफल) प्रयोग करें।
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।
The sum is \(2-\sqrt{5}\), so the coefficient of (x) is (-\(2-\sqrt{5}\)=\sqrt{5}-2). The product \(-2\sqrt{5}\) also matches.
Step 2
Why this answer is correct
The correct answer is A. (2) और \(-\sqrt{5}\) / (2) and \(-\sqrt{5}\). The sum is \(2-\sqrt{5}\), so the coefficient of (x) is (-\(2-\sqrt{5}\)=\sqrt{5}-2). The product \(-2\sqrt{5}\) also matches.
Step 3
Exam Tip
योग \(2-\sqrt{5}\) है, इसलिए (x) का गुणांक (-\(2-\sqrt{5}\)=\sqrt{5}-2) है। गुणनफल \(-2\sqrt{5}\) भी सही है।
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।
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।
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।
The zeroes are \(x=\pm\sqrt{2}\), and \(\sqrt{2}\) is irrational. In exams, simplify square-root zeroes before deciding the type.
Step 2
Why this answer is correct
The correct answer is B. दोनों अपरिमेय हैं / Both are irrational. The zeroes are \(x=\pm\sqrt{2}\), and \(\sqrt{2}\) is irrational. In exams, simplify square-root zeroes before deciding the type.
Step 3
Exam Tip
शून्यक \(x=\pm\sqrt{2}\) हैं और \(\sqrt{2}\) अपरिमेय है। परीक्षा में वर्गमूल वाले शून्यकों को सरल करके जाँचें।
The original zeroes are (2) and (4), so the new zeroes are (4) and (16). The new polynomial is \(x^2-20x+64\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2-20x+64\). The original zeroes are (2) and (4), so the new zeroes are (4) and (16). The new polynomial is \(x^2-20x+64\).
Step 3
Exam Tip
मूल शून्यक (2) और (4) हैं, इसलिए नए शून्यक (4) और (16) हैं। नया बहुपद \(x^2-20x+64\) है।
A. दिए गए आधार पर कोई शून्यक नहीं दिखता/No zero is shown from the given data
Step 1
Concept
Zeroes are linked only with the (x)-axis where (y=0). Intersections with (y=2) do not show zeroes.
Step 2
Why this answer is correct
The correct answer is A. दिए गए आधार पर कोई शून्यक नहीं दिखता / No zero is shown from the given data. Zeroes are linked only with the (x)-axis where (y=0). Intersections with (y=2) do not show zeroes.
Step 3
Exam Tip
शून्यक केवल (x)-अक्ष यानी (y=0) से जुड़े होते हैं। (y=2) से प्रतिच्छेद शून्यक नहीं बताता।
A. इससे शून्यक निश्चित नहीं होता/A zero cannot be determined from this alone
Step 1
Concept
The (y)-intercept tells (p(0)) not all zeroes. Zeroes need (x)-axis intersections.
Step 2
Why this answer is correct
The correct answer is A. इससे शून्यक निश्चित नहीं होता / A zero cannot be determined from this alone. The (y)-intercept tells (p(0)) not all zeroes. Zeroes need (x)-axis intersections.
Step 3
Exam Tip
(y)-प्रतिच्छेद (p(0)) बताता है न कि सभी शून्यक। शून्यक के लिए (x)-अक्ष से प्रतिच्छेद चाहिए।
A. दो भिन्न वास्तविक शून्यक/Two distinct real zeroes
Step 1
Concept
Two separate intersections give two distinct real zeroes. Different (x)-intercepts mean different zeroes.
Step 2
Why this answer is correct
The correct answer is A. दो भिन्न वास्तविक शून्यक / Two distinct real zeroes. Two separate intersections give two distinct real zeroes. Different (x)-intercepts mean different zeroes.
Step 3
Exam Tip
दो अलग कटान दो अलग वास्तविक शून्यक देते हैं। ग्राफ में अलग (x)-प्रतिच्छेद अलग शून्यक होते हैं।
It is ((x-d)2-36), so \(x-d=\pm6\) and the zeroes are (d-6), (d+6). Tip: use difference of squares.
Step 2
Why this answer is correct
The correct answer is A. (d-6) और (d+6) / (d-6) and (d+6). It is ((x-d)2-36), so \(x-d=\pm6\) and the zeroes are (d-6), (d+6). Tip: use difference of squares.
Step 3
Exam Tip
यह ((x-d)2-36) है, इसलिए \(x-d=\pm6\) और शून्यक (d-6), (d+6) हैं। टिप: वर्गों के अंतर का उपयोग करें।
A. (22) को (10) करना होगा/(22) must be changed to (10)
Step 1
Concept
For equal distance from the (y)-axis, zeroes should be opposites, so (10) is needed with (-10). Tip: symmetric zeroes are (a) and (-a).
Step 2
Why this answer is correct
The correct answer is A. (22) को (10) करना होगा / (22) must be changed to (10). For equal distance from the (y)-axis, zeroes should be opposites, so (10) is needed with (-10). Tip: symmetric zeroes are (a) and (-a).
Step 3
Exam Tip
(y)-अक्ष से समान दूरी के लिए शून्यक विपरीत होने चाहिए, इसलिए (-10) के साथ (10) चाहिए। टिप: सममित शून्यक (a) और (-a) होते हैं।
For a downward-opening parabola, values between the two zeroes are positive. Tip: (x=2) lies between the zeroes.
Step 2
Why this answer is correct
The correct answer is A. (x)-अक्ष के ऊपर / Above the (x)-axis. For a downward-opening parabola, values between the two zeroes are positive. Tip: (x=2) lies between the zeroes.
Step 3
Exam Tip
नीचे खुलने वाले परवलय में दो शून्यकों के बीच मान धनात्मक होते हैं। टिप: (x=2) दोनों शून्यकों के बीच है।
The zeroes are (-12) and (12), so the product is (-144) and the sum is (0). Tip: opposite zeroes have sum (0).
Step 2
Why this answer is correct
The correct answer is A. गुणनफल (-144), योग (0) / Product (-144), sum (0). The zeroes are (-12) and (12), so the product is (-144) and the sum is (0). Tip: opposite zeroes have sum (0).
Step 3
Exam Tip
शून्यक (-12) और (12) हैं, इसलिए गुणनफल (-144) और योग (0) है। टिप: विपरीत शून्यकों का योग (0) होता है।
The axis of symmetry is at the average of the zeroes, (\frac{(q-11)+(q+7)}{2}=q-2). Tip: take the midpoint even with symbols.
Step 2
Why this answer is correct
The correct answer is A. (x=q-2). The axis of symmetry is at the average of the zeroes, (\frac{(q-11)+(q+7)}{2}=q-2). Tip: take the midpoint even with symbols.
Step 3
Exam Tip
सममिति अक्ष शून्यकों के औसत पर है, (\frac{(q-11)+(q+7)}{2}=q-2)। टिप: प्रतीकों में भी मध्य मान लें।
The origin is also on the (x)-axis, and (x=-9) is another (x)-axis intersection. Tip: count ((0,0)) as zero (0).
Step 2
Why this answer is correct
The correct answer is A. (0) और (-9) / (0) and (-9). The origin is also on the (x)-axis, and (x=-9) is another (x)-axis intersection. Tip: count ((0,0)) as zero (0).
Step 3
Exam Tip
मूल बिंदु (x)-अक्ष पर भी है और (x=-9) भी (x)-अक्ष कटान है। टिप: ((0,0)) को शून्यक (0) के रूप में गिनें।
From (x-c=0) we get (c), and from (x+d=0) we get (-d). Tip: do not count repetition among distinct zeroes.
Step 2
Why this answer is correct
The correct answer is A. (c) और (-d) / (c) and (-d). From (x-c=0) we get (c), and from (x+d=0) we get (-d). Tip: do not count repetition among distinct zeroes.
Step 3
Exam Tip
(x-c=0) से (c) और (x+d=0) से (-d) मिलता है। टिप: अलग शून्यकों में दोहराव न गिनें।
There are two distinct zeroes (-2) and (5), and both have even powers. Tip: at an even-power zero the graph usually touches.
Step 2
Why this answer is correct
The correct answer is A. दो, दोनों पर स्पर्श / Two, touches at both. There are two distinct zeroes (-2) and (5), and both have even powers. Tip: at an even-power zero the graph usually touches.
Step 3
Exam Tip
दो अलग शून्यक (-2) और (5) हैं तथा दोनों की घात सम है। टिप: सम घात वाले शून्यक पर ग्राफ सामान्यतः स्पर्श करता है।
(x=-4) lies between the two zeroes and an upward-opening parabola stays below there. Tip: check the sign region between zeroes.
Step 2
Why this answer is correct
The correct answer is A. (x)-अक्ष के नीचे / Below the (x)-axis. (x=-4) lies between the two zeroes and an upward-opening parabola stays below there. Tip: check the sign region between zeroes.
Step 3
Exam Tip
(x=-4) दोनों शून्यकों के बीच है और ऊपर खुलने वाला परवलय बीच में नीचे रहता है। टिप: शून्यकों के बीच संकेत क्षेत्र देखें।
It is ((x-c)2-25), so \(x-c=\pm5\) and the zeroes are (c-5), (c+5). Tip: use difference of squares.
Step 2
Why this answer is correct
The correct answer is A. (c-5) और (c+5) / (c-5) and (c+5). It is ((x-c)2-25), so \(x-c=\pm5\) and the zeroes are (c-5), (c+5). Tip: use difference of squares.
Step 3
Exam Tip
यह ((x-c)2-25) है, इसलिए \(x-c=\pm5\) और शून्यक (c-5), (c+5) हैं। टिप: वर्गों के अंतर का उपयोग करें।
A. (18) को (8) करना होगा/(18) must be changed to (8)
Step 1
Concept
For equal distance from the (y)-axis, zeroes should be opposites, so (8) is needed with (-8). Tip: symmetric zeroes are (a) and (-a).
Step 2
Why this answer is correct
The correct answer is A. (18) को (8) करना होगा / (18) must be changed to (8). For equal distance from the (y)-axis, zeroes should be opposites, so (8) is needed with (-8). Tip: symmetric zeroes are (a) and (-a).
Step 3
Exam Tip
(y)-अक्ष से समान दूरी के लिए शून्यक विपरीत होने चाहिए, इसलिए (-8) के साथ (8) चाहिए। टिप: सममित शून्यक (a) और (-a) होते हैं।
