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।
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।
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।
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।
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 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।
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।
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. 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।
A. जो (x)-अक्ष को दो अलग बिंदुओं पर काटे/One that cuts the (x)-axis at two distinct points
Step 1
Concept
Two real zeroes need two distinct (x)-axis intersections. Tip: (y)-axis intersections do not count as zeroes.
Step 2
Why this answer is correct
The correct answer is A. जो (x)-अक्ष को दो अलग बिंदुओं पर काटे / One that cuts the (x)-axis at two distinct points. Two real zeroes need two distinct (x)-axis intersections. Tip: (y)-axis intersections do not count as zeroes.
Step 3
Exam Tip
दो वास्तविक शून्यक के लिए दो अलग (x)-अक्ष कटान चाहिए। टिप: (y)-अक्ष कटान शून्यक नहीं गिनाता।
Both cutting and touching mean meeting the (x)-axis. There are two distinct points, so there are two distinct real zeroes.
Step 2
Why this answer is correct
The correct answer is A. दो / Two. Both cutting and touching mean meeting the (x)-axis. There are two distinct points, so there are two distinct real zeroes.
Step 3
Exam Tip
कटना और छूना दोनों (x)-अक्ष से मिलना है। दो अलग बिंदु हैं, इसलिए दो अलग वास्तविक शून्यक हैं।
A. दो अलग-अलग बिंदुओं पर काटेगा/It will cut at two distinct points
Step 1
Concept
Two real zeroes show two distinct (x)-axis intersections. Hence the parabola cuts the (x)-axis at two points.
Step 2
Why this answer is correct
The correct answer is A. दो अलग-अलग बिंदुओं पर काटेगा / It will cut at two distinct points. Two real zeroes show two distinct (x)-axis intersections. Hence the parabola cuts the (x)-axis at two points.
Step 3
Exam Tip
दो वास्तविक शून्यक दो अलग (x)-अक्ष कटाव बताते हैं। इसलिए परवलय (x)-अक्ष को दो बिंदुओं पर काटेगा।
Each distinct intersection with the (x)-axis gives one real zero. Therefore three distinct intersections give three real zeroes.
Step 2
Why this answer is correct
The correct answer is A. तीन / Three. Each distinct intersection with the (x)-axis gives one real zero. Therefore three distinct intersections give three real zeroes.
Step 3
Exam Tip
हर अलग (x)-अक्ष कटाव एक वास्तविक शून्यक देता है। इसलिए तीन अलग कटावों से तीन वास्तविक शून्यक मिलेंगे।
Each distinct intersection with the (x)-axis gives one real zero. With two intersections, there are two real zeroes.
Step 2
Why this answer is correct
The correct answer is A. दो / Two. Each distinct intersection with the (x)-axis gives one real zero. With two intersections, there are two real zeroes.
Step 3
Exam Tip
(x)-अक्ष से प्रत्येक अलग कटाव एक वास्तविक शून्यक देता है। दो कटाव होने पर दो वास्तविक शून्यक होंगे।
(4) is rational and \(\sqrt{13}\) is irrational, so the sum is irrational. In exams identify square roots of perfect squares first.
Step 2
Why this answer is correct
The correct answer is A. \(4+\sqrt{13}\). (4) is rational and \(\sqrt{13}\) is irrational, so the sum is irrational. In exams identify square roots of perfect squares first.
Step 3
Exam Tip
(4) परिमेय है और \(\sqrt{13}\) अपरिमेय है, इसलिए योग अपरिमेय है। परीक्षा में पूर्ण वर्ग के वर्गमूल को पहले पहचानें।
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. \(\sqrt{2}\) 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।
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।
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।
There are two distinct zeroes (1) and (-4), 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 (1) and (-4), and both have even powers. Tip: at an even-power zero the graph usually touches.
Step 3
Exam Tip
दो अलग शून्यक (1) और (-4) हैं तथा दोनों की घात सम है। टिप: सम घात वाले शून्यक पर ग्राफ सामान्यतः स्पर्श करता है।
(x-2+x-6=(x+3)(x-2)), so the zeroes are (-3) and (2). Tip: read the signs in the factors carefully.
Step 2
Why this answer is correct
The correct answer is A. (2) और (-3) / (2) and (-3). (x-2+x-6=(x+3)(x-2)), so the zeroes are (-3) and (2). Tip: read the signs in the factors carefully.
Step 3
Exam Tip
(x-2+x-6=(x+3)(x-2)) इसलिए शून्यक (-3) और (2) हैं। टिप: गुणनखंडों के चिह्न ध्यान से पढ़ें।
A. जो (x)-अक्ष को दो अलग बिंदुओं पर काटे/One that cuts the (x)-axis at two distinct points
Step 1
Concept
Two distinct (x)-axis intersections give two distinct real zeroes. Tip: do not count (y)-axis intersections as zeroes.
Step 2
Why this answer is correct
The correct answer is A. जो (x)-अक्ष को दो अलग बिंदुओं पर काटे / One that cuts the (x)-axis at two distinct points. Two distinct (x)-axis intersections give two distinct real zeroes. Tip: do not count (y)-axis intersections as zeroes.
Step 3
Exam Tip
दो अलग (x)-अक्ष कटान दो अलग वास्तविक शून्यक देते हैं। टिप: (y)-अक्ष कटान को शून्यक न गिनें।
There are three distinct (x)-axis intersections, so there are three real zeroes. Tip: count only points with (y=0).
Step 2
Why this answer is correct
The correct answer is C. तीन / Three. There are three distinct (x)-axis intersections, so there are three real zeroes. Tip: count only points with (y=0).
Step 3
Exam Tip
तीन अलग (x)-अक्ष कटान हैं इसलिए तीन वास्तविक शून्यक हैं। टिप: केवल (y=0) वाले बिंदु गिनें।
A zero (a) gives the intersection point ((a,0)). Tip: do not make the zero the (y)-coordinate.
Step 2
Why this answer is correct
The correct answer is A. ((1,0)) और ((4,0)) / ((1,0)) and ((4,0)). A zero (a) gives the intersection point ((a,0)). Tip: do not make the zero the (y)-coordinate.
Step 3
Exam Tip
शून्यक (a) से कटान बिंदु ((a,0)) बनता है। टिप: शून्यक को (y)-निर्देशांक न बनाएं।
A. जब उसका ग्राफ (x)-अक्ष को दो अलग बिंदुओं पर काटे/When its graph cuts the (x)-axis at two distinct points
Step 1
Concept
Real zeroes of a quadratic polynomial are found from points where it meets the (x)-axis. Two distinct intersections give two real zeroes.
Step 2
Why this answer is correct
The correct answer is A. जब उसका ग्राफ (x)-अक्ष को दो अलग बिंदुओं पर काटे / When its graph cuts the (x)-axis at two distinct points. Real zeroes of a quadratic polynomial are found from points where it meets the (x)-axis. Two distinct intersections give two real zeroes.
Step 3
Exam Tip
द्विघात बहुपद के वास्तविक शून्यक (x)-अक्ष से मिलने वाले बिंदुओं से मिलते हैं। दो अलग कटाव दो वास्तविक शून्यक देते हैं।
