A. \(\sqrt{2}=-\frac{p}{q}\), इसलिए \(\sqrt{2}\) परिमेय होगा जो असंभव है/\(\sqrt{2}=-\frac{p}{q}\), so \(\sqrt{2}\) would be rational which is impossible
Step 1
Concept
Since \(-\frac{p}{q}\) is rational, this would make \(\sqrt{2}\) rational which is false. In exams recognize the contradiction method.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}=-\frac{p}{q}\), इसलिए \(\sqrt{2}\) परिमेय होगा जो असंभव है / \(\sqrt{2}=-\frac{p}{q}\), so \(\sqrt{2}\) would be rational which is impossible. Since \(-\frac{p}{q}\) is rational, this would make \(\sqrt{2}\) rational which is false. In exams recognize the contradiction method.
Step 3
Exam Tip
क्योंकि \(-\frac{p}{q}\) परिमेय है, इससे \(\sqrt{2}\) परिमेय मानना पड़ेगा जो गलत है। परीक्षा में विरोधाभास विधि को पहचानें।
B. यह अनवसानी आवर्ती दशमलव है/It is non-terminating recurring decimal
Step 1
Concept
The denominator contains (13), and after simplification the denominator is not made only of (2) and (5). In exams always check prime factors of the denominator.
Step 2
Why this answer is correct
The correct answer is B. यह अनवसानी आवर्ती दशमलव है / It is non-terminating recurring decimal. The denominator contains (13), and after simplification the denominator is not made only of (2) and (5). In exams always check prime factors of the denominator.
Step 3
Exam Tip
हर में (13) है और भिन्न सरल करने पर भी केवल (2) और (5) नहीं बचते। परीक्षा में हर के अभाज्य गुणनखंड जरूर जांचें।
A. \(8+2\sqrt{15}\), अपरिमेय/\(8+2\sqrt{15}\), irrational
Step 1
Concept
\(x^2=5+3+2\sqrt{15}=8+2\sqrt{15}\), so it is irrational. In exams use ((a+b)2) carefully.
Step 2
Why this answer is correct
The correct answer is A. \(8+2\sqrt{15}\), अपरिमेय / \(8+2\sqrt{15}\), irrational. \(x^2=5+3+2\sqrt{15}=8+2\sqrt{15}\), so it is irrational. In exams use ((a+b)2) carefully.
Step 3
Exam Tip
\(x^2=5+3+2\sqrt{15}=8+2\sqrt{15}\), इसलिए यह अपरिमेय है। परीक्षा में ((a+b)2) का प्रयोग सावधानी से करें।
\(\sqrt{45}=3\sqrt{5}\), which is real and irrational. In exams do not treat the square root of a negative number as real.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{45}\). \(\sqrt{45}=3\sqrt{5}\), which is real and irrational. In exams do not treat the square root of a negative number as real.
Step 3
Exam Tip
\(\sqrt{45}=3\sqrt{5}\), जो वास्तविक और अपरिमेय है। परीक्षा में ऋणात्मक वर्गमूल को वास्तविक संख्या न मानें।
(\frac{2}{\sqrt{3}+1}\times\frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{2\(\sqrt{3}-1\)}{2}=\sqrt{3}-1). The conjugate makes the denominator rational.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{3}-1\). (\frac{2}{\sqrt{3}+1}\times\frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{2\(\sqrt{3}-1\)}{2}=\sqrt{3}-1). The conjugate makes the denominator rational.
Step 3
Exam Tip
(\frac{2}{\sqrt{3}+1}\times\frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{2\(\sqrt{3}-1\)}{2}=\sqrt{3}-1) है। परीक्षा में संयुग्मी से हर परिमेय बनता है।
\(\frac{1}{\sqrt{5}-2}\times\frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2\). Rationalise the denominator in exams.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{5}+2\). \(\frac{1}{\sqrt{5}-2}\times\frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2\). Rationalise the denominator in exams.
Step 3
Exam Tip
\(\frac{1}{\sqrt{5}-2}\times\frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2\) है। परीक्षा में हर का परिमेयकरण करें।
\(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\), so the difference is \(\sqrt{3}\). Simplify first in exams.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{3}\). \(\sqrt{27}=3\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\), so the difference is \(\sqrt{3}\). Simplify first in exams.
Step 3
Exam Tip
\(\sqrt{27}=3\sqrt{3}\) और \(\sqrt{12}=2\sqrt{3}\), इसलिए अंतर \(\sqrt{3}\) है। परीक्षा में पहले सरलीकरण करें।
A. \(3\sqrt{2}\), अपरिमेय/\(3\sqrt{2}\), irrational
Step 1
Concept
\(\sqrt{8}=2\sqrt{2}\), so \(a=3\sqrt{2}\), irrational. Combine like radicals in exams.
Step 2
Why this answer is correct
The correct answer is A. \(3\sqrt{2}\), अपरिमेय / \(3\sqrt{2}\), irrational. \(\sqrt{8}=2\sqrt{2}\), so \(a=3\sqrt{2}\), irrational. Combine like radicals in exams.
Step 3
Exam Tip
\(\sqrt{8}=2\sqrt{2}\), इसलिए \(a=3\sqrt{2}\) अपरिमेय है। परीक्षा में समान करणी वाले पद जोड़ें।
Square roots of distinct primes are different irrationals and their sum cannot be rational. In exams do not assume independent radicals can combine to a rational number.
Step 2
Why this answer is correct
The correct answer is B. यह असंभव है / This is impossible. Square roots of distinct primes are different irrationals and their sum cannot be rational. In exams do not assume independent radicals can combine to a rational number.
Step 3
Exam Tip
अलग अभाज्य संख्याओं के वर्गमूल अलग अपरिमेय होते हैं और उनका योग परिमेय नहीं हो सकता। परीक्षा में स्वतंत्र वर्गमूलों को जोड़कर परिमेय न मानें।
For a positive integer (m), \(\sqrt{m}\) is rational only when (m) is a perfect square. Identifying perfect squares is important in exams.
Step 2
Why this answer is correct
The correct answer is B. (m) पूर्ण वर्ग है / (m) is a perfect square. For a positive integer (m), \(\sqrt{m}\) is rational only when (m) is a perfect square. Identifying perfect squares is important in exams.
Step 3
Exam Tip
धनात्मक पूर्णांक (m) के लिए \(\sqrt{m}\) परिमेय तभी होगा जब (m) पूर्ण वर्ग हो। परीक्षा में पूर्ण वर्ग पहचानना जरूरी है।
C. सिर्फ \(\sqrt{2}\) ही अकेला अपरिमेय शून्यक हो और गुणांक परिमेय रहें/Only \(\sqrt{2}\) is the sole irrational zero while coefficients stay rational
Step 1
Concept
In a quadratic with rational coefficients an irrational zero comes with its conjugate. In exams be suspicious of a lone irrational root.
Step 2
Why this answer is correct
The correct answer is C. सिर्फ \(\sqrt{2}\) ही अकेला अपरिमेय शून्यक हो और गुणांक परिमेय रहें / Only \(\sqrt{2}\) is the sole irrational zero while coefficients stay rational. In a quadratic with rational coefficients an irrational zero comes with its conjugate. In exams be suspicious of a lone irrational root.
Step 3
Exam Tip
परिमेय गुणांकों वाले द्विघात में अपरिमेय शून्यक अपने संयुग्मी के साथ आता है। परीक्षा में अकेले अपरिमेय मूल पर संदेह करें।
Here \(x^2=5+2\sqrt{6}\) and then (\(x^2-5\)2=24), so \(x^4-10x^2+1=0\). In exams a sum of two radicals may lead to a fourth-degree relation.
Step 2
Why this answer is correct
The correct answer is C. \(x^4-10x^2+1=0\). Here \(x^2=5+2\sqrt{6}\) and then (\(x^2-5\)2=24), so \(x^4-10x^2+1=0\). In exams a sum of two radicals may lead to a fourth-degree relation.
Step 3
Exam Tip
\(x^2=5+2\sqrt{6}\) और फिर (\(x^2-5\)2=24), इसलिए \(x^4-10x^2+1=0\) है। परीक्षा में दो वर्गमूलों के योग से कभी द्विघात नहीं बल्कि चतुर्थ घात संबंध भी बन सकता है।
(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।
A. वास्तविक गुणांकों वाला द्विघात बहुपद/Quadratic polynomial with real coefficients
Step 1
Concept
\(\sqrt{2}\) is real but irrational, so the coefficients are real but not all rational. Since the degree is (2), it is quadratic.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक गुणांकों वाला द्विघात बहुपद / Quadratic polynomial with real coefficients. \(\sqrt{2}\) is real but irrational, so the coefficients are real but not all rational. Since the degree is (2), it is quadratic.
Step 3
Exam Tip
\(\sqrt{2}\) वास्तविक लेकिन अपरिमेय है, इसलिए गुणांक वास्तविक हैं पर सभी परिमेय नहीं। घात (2) होने से यह द्विघात है।
For (k=2), the discriminant is (16-8=8), positive but not a perfect square. Therefore the roots are real and irrational.
Step 2
Why this answer is correct
The correct answer is B. (k=2). For (k=2), the discriminant is (16-8=8), positive but not a perfect square. Therefore the roots are real and irrational.
Step 3
Exam Tip
(k=2) पर विविक्तकर (16-8=8), जो धनात्मक पर पूर्ण वर्ग नहीं है। इसलिए मूल वास्तविक और अपरिमेय होंगे।
C. कोई वास्तविक मूल नहीं है/There are no real roots
Step 1
Concept
The discriminant is (4-20=-16), which is negative, so there are no real roots. In exams do not treat a negative discriminant as real zeroes.
Step 2
Why this answer is correct
The correct answer is C. कोई वास्तविक मूल नहीं है / There are no real roots. The discriminant is (4-20=-16), which is negative, so there are no real roots. In exams do not treat a negative discriminant as real zeroes.
Step 3
Exam Tip
विविक्तकर (4-20=-16) ऋणात्मक है, इसलिए वास्तविक मूल नहीं हैं। परीक्षा में ऋणात्मक विविक्तकर को वास्तविक शून्यक नहीं मानें।
The discriminant is (16-16=0), so the roots are equal and (x=2) is rational. In exams (D=0) means equal roots.
Step 2
Why this answer is correct
The correct answer is B. दो समान परिमेय / Two equal rational roots. The discriminant is (16-16=0), so the roots are equal and (x=2) is rational. In exams (D=0) means equal roots.
Step 3
Exam Tip
विविक्तकर (16-16=0), इसलिए मूल समान हैं और (x=2) परिमेय है। परीक्षा में (D=0) का अर्थ समान मूल होता है।
C. दो अलग अपरिमेय वास्तविक/Two distinct irrational real values
Step 1
Concept
The discriminant is (36-4=32), positive but not a perfect square. So there are two distinct irrational real roots.
Step 2
Why this answer is correct
The correct answer is C. दो अलग अपरिमेय वास्तविक / Two distinct irrational real values. The discriminant is (36-4=32), positive but not a perfect square. So there are two distinct irrational real roots.
Step 3
Exam Tip
विविक्तकर (36-4=32) है, जो पूर्ण वर्ग नहीं है और धनात्मक है। इसलिए दो अलग अपरिमेय वास्तविक मूल मिलते हैं।
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।
Because (\(3+\sqrt{8}\)\(3-\sqrt{8}\)=9-8=1), the reciprocal is \(3-\sqrt{8}\). If the product is (1), the reciprocal is immediate.
Step 2
Why this answer is correct
The correct answer is A. \(3-\sqrt{8}\). Because (\(3+\sqrt{8}\)\(3-\sqrt{8}\)=9-8=1), the reciprocal is \(3-\sqrt{8}\). If the product is (1), the reciprocal is immediate.
Step 3
Exam Tip
क्योंकि (\(3+\sqrt{8}\)\(3-\sqrt{8}\)=9-8=1), इसलिए व्युत्क्रम \(3-\sqrt{8}\) है। परीक्षा में गुणनफल (1) होने पर व्युत्क्रम तुरंत मिल जाता है।
\(\frac{1}{1+\sqrt{2}}=\sqrt{2}-1\), so the sum is \(2\sqrt{2}\). Rationalising the denominator is a quick exam method.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{2}\). \(\frac{1}{1+\sqrt{2}}=\sqrt{2}-1\), so the sum is \(2\sqrt{2}\). Rationalising the denominator is a quick exam method.
Step 3
Exam Tip
\(\frac{1}{1+\sqrt{2}}=\sqrt{2}-1\), इसलिए योग \(2\sqrt{2}\) है। परीक्षा में हर का परिमेयकरण तेज तरीका है।
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।
For a quadratic with rational coefficients, if \(a+\sqrt{b}\) is a zero then \(a-\sqrt{b}\) is also a zero. The conjugate-root rule is useful in exams.
Step 2
Why this answer is correct
The correct answer is A. \(2-\sqrt{3}\). For a quadratic with rational coefficients, if \(a+\sqrt{b}\) is a zero then \(a-\sqrt{b}\) is also a zero. The conjugate-root rule is useful in exams.
Step 3
Exam Tip
परिमेय गुणांकों वाले द्विघात में \(a+\sqrt{b}\) के साथ \(a-\sqrt{b}\) भी शून्यक होता है। परीक्षा में संयुग्मी मूल का नियम उपयोगी है।
(p\(\sqrt{2}\)=2-5\sqrt{2}+6=8-5\sqrt{2}), which is irrational. In exams substitute first and then simplify the radical.
Step 2
Why this answer is correct
The correct answer is B. अपरिमेय / Irrational. (p\(\sqrt{2}\)=2-5\sqrt{2}+6=8-5\sqrt{2}), which is irrational. In exams substitute first and then simplify the radical.
Step 3
Exam Tip
(p\(\sqrt{2}\)=2-5\sqrt{2}+6=8-5\sqrt{2}), जो अपरिमेय है। परीक्षा में पहले प्रतिस्थापन करें फिर वर्गमूल को सरल करें।
Since \(\sqrt{8}\times\sqrt{18}=\sqrt{144}=12\), it is rational. In exams simplify products of radicals first.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{8}\times\sqrt{18}\). Since \(\sqrt{8}\times\sqrt{18}=\sqrt{144}=12\), it is rational. In exams simplify products of radicals first.
Step 3
Exam Tip
\(\sqrt{8}\times\sqrt{18}=\sqrt{144}=12\), इसलिए यह परिमेय है। परीक्षा में गुणन में वर्गमूलों को पहले एक साथ सरल करें।
Here \(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\), so the difference is \(3\sqrt{2}\), irrational; no listed expression is rational, so this item must be checked carefully.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{50}-\sqrt{8}\). Here \(\sqrt{50}=5\sqrt{2}\) and \(\sqrt{8}=2\sqrt{2}\), so the difference is \(3\sqrt{2}\), irrational; no listed expression is rational, so this item must be checked carefully.
Step 3
Exam Tip
\(\sqrt{50}=5\sqrt{2}\) और \(\sqrt{8}=2\sqrt{2}\), इसलिए अंतर \(3\sqrt{2}\) अपरिमेय है; सही परिमेय विकल्प नहीं दिखता, अतः ध्यान दें कि \(\sqrt{7}\sqrt{14}=7\sqrt{2}\) भी अपरिमेय है।
The product of a non-zero rational number and an irrational number remains irrational. In exams the condition \(a\ne0\) is very important.
Step 2
Why this answer is correct
The correct answer is B. हमेशा अपरिमेय / Always irrational. The product of a non-zero rational number and an irrational number remains irrational. In exams the condition \(a\ne0\) is very important.
Step 3
Exam Tip
अशून्य परिमेय संख्या से अपरिमेय संख्या का गुणन अपरिमेय रहता है। परीक्षा में \(a\ne0\) शर्त बहुत महत्वपूर्ण है।
If (x) were rational then \(\sqrt{2}+x\) would be irrational. So (x) must be irrational; remember the sum rule for rational and irrational numbers.
Step 2
Why this answer is correct
The correct answer is B. (x) अपरिमेय है / (x) is irrational. If (x) were rational then \(\sqrt{2}+x\) would be irrational. So (x) must be irrational; remember the sum rule for rational and irrational numbers.
Step 3
Exam Tip
यदि (x) परिमेय होता तो \(\sqrt{2}+x\) अपरिमेय होता। इसलिए (x) अपरिमेय होना चाहिए; परीक्षा में परिमेय और अपरिमेय के योग का नियम याद रखें।
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।
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।
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।
(\(1+\sqrt{3}\)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।
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।
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. 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।
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. (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।
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. 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।
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. 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. \(-\sqrt{7}+1\) 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 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. 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 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।
\(\alpha+\beta=2\) 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।
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।
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।
(\(\alpha-\beta\)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. \(-2+\sqrt{2}\) 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।
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।
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।
\(\alpha+\beta=2\) 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।
