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।
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।
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।
If \(\frac{s}{r}\) were rational then \(s=r\cdot\frac{s}{r}\) would be rational which is false. In exams check the non-zero condition.
Step 2
Why this answer is correct
The correct answer is A. अपरिमेय / Irrational. If \(\frac{s}{r}\) were rational then \(s=r\cdot\frac{s}{r}\) would be rational which is false. In exams check the non-zero condition.
Step 3
Exam Tip
यदि \(\frac{s}{r}\) परिमेय हो तो \(s=r\cdot\frac{s}{r}\) परिमेय हो जाएगा जो गलत है। परीक्षा में शून्येतर शर्त जरूर देखें।
A non zero rational multiplier keeps an irrational number irrational. The condition \(r\neq0\) is important.
Step 2
Why this answer is correct
The correct answer is A. अपरिमेय संख्या / Irrational number. A non zero rational multiplier keeps an irrational number irrational. The condition \(r\neq0\) is important.
Step 3
Exam Tip
गैर शून्य परिमेय गुणक अपरिमेय संख्या को अपरिमेय ही रखता है। शर्त \(r\neq0\) महत्वपूर्ण है।
Dividing by a non-zero rational number is the same as multiplying by its reciprocal.
Step 2
Why this answer is correct
An irrational number multiplied by a non-zero rational number remains irrational.
Step 3
Exam Tip
Convert division questions into multiplication for easier reasoning. चरण 1: शून्य रहित परिमेय से भाग देना उसी के व्युत्क्रम से गुणा करना है। चरण 2: अपरिमेय संख्या को शून्य रहित परिमेय से गुणा करने पर अपरिमेय संख्या मिलती है। चरण 3: भाग के प्रश्नों को गुणा में बदलकर सोचें।
Dividing by a non-zero rational number is the same as multiplying by its reciprocal.
Step 2
Why this answer is correct
\(\frac{1}{r}\) is also a non-zero rational number, so \(\frac{x}{r}\) remains irrational.
Step 3
Exam Tip
In division questions, always check that the denominator is not zero. चरण 1: अशून्य परिमेय संख्या से भाग देना उसी के व्युत्क्रम से गुणा करने जैसा है। चरण 2: \(\frac{1}{r}\) भी अशून्य परिमेय है, इसलिए \(\frac{x}{r}\) अपरिमेय रहेगा। चरण 3: भाग वाले प्रश्न में हर के शून्य न होने की शर्त जरूर देखें।
Multiplying an irrational number by a non-zero rational number keeps it irrational.
Step 2
Why this answer is correct
If (pq) were rational, then \(q=\frac{pq}{p}\) would be rational, which contradicts the given condition.
Step 3
Exam Tip
Always check that the rational multiplier is not zero. चरण 1: अशून्य परिमेय संख्या से गुणा करने पर अपरिमेयता बनी रहती है। चरण 2: यदि (pq) परिमेय मान लें, तो \(q=\frac{pq}{p}\) परिमेय हो जाएगा, जो गलत है। चरण 3: परीक्षा में ध्यान रखें कि (p) शून्य नहीं होना चाहिए।
Dividing by a non-zero rational number does not remove irrationality.
Step 2
Why this answer is correct
For example, \(\frac{\sqrt{3}}{4}\) remains irrational.
Step 3
Exam Tip
The condition \(k\neq0\) is necessary because division by zero is not possible. चरण 1: अशून्य परिमेय संख्या से भाग करने पर अपरिमेयता समाप्त नहीं होती। चरण 2: जैसे \(\frac{\sqrt{3}}{4}\) अपरिमेय रहता है। चरण 3: यहां \(k\neq0\) जरूरी है क्योंकि शून्य से भाग संभव नहीं है।
Dividing by a non-zero rational number does not remove irrationality.
Step 2
Why this answer is correct
For example, \(\frac{\sqrt{3}}{4}\) remains irrational.
Step 3
Exam Tip
The condition \(k\neq0\) is necessary because division by zero is not possible. चरण 1: अशून्य परिमेय संख्या से भाग करने पर अपरिमेयता समाप्त नहीं होती। चरण 2: जैसे \(\frac{\sqrt{3}}{4}\) अपरिमेय रहता है। चरण 3: यहां \(k\neq0\) जरूरी है क्योंकि शून्य से भाग संभव नहीं है।
A non-zero rational multiplier does not remove irrationality.
Step 2
Why this answer is correct
For example, \(4\sqrt{3}\) remains irrational.
Step 3
Exam Tip
The non-zero condition is important because multiplying by (0) gives (0). चरण 1: अशून्य परिमेय गुणक अपरिमेयता को समाप्त नहीं करता। चरण 2: जैसे \(4\sqrt{3}\) अपरिमेय रहता है। चरण 3: यहां अशून्य शर्त जरूरी है क्योंकि (0) से गुणा करने पर परिणाम (0) होगा।
Dividing by a non-zero rational number does not remove irrationality.
Step 2
Why this answer is correct
For example, \(\frac{\sqrt{2}}{5}\) is irrational.
Step 3
Exam Tip
The condition \(r\neq0\) is necessary because division by zero is not possible. चरण 1: अशून्य परिमेय संख्या से भाग करने पर अपरिमेयता समाप्त नहीं होती। चरण 2: जैसे \(\frac{\sqrt{2}}{5}\) अपरिमेय है। चरण 3: यहां \(r\neq0\) शर्त जरूरी है क्योंकि शून्य से भाग संभव नहीं होता।
A non-zero rational multiplier does not remove irrationality.
Step 2
Why this answer is correct
For example, \(3\sqrt{2}\) is irrational.
Step 3
Exam Tip
The non-zero condition is important because multiplying by zero gives zero. चरण 1: अशून्य परिमेय गुणक अपरिमेयता को समाप्त नहीं करता। चरण 2: जैसे \(3\sqrt{2}\) अपरिमेय है। चरण 3: यहां अशून्य शर्त जरूरी है क्योंकि शून्य से गुणा करने पर परिणाम शून्य होगा।
Multiplying an irrational number by a non-zero rational number keeps it irrational.
Step 2
Why this answer is correct
For example, \(2 \times \sqrt{3}=2\sqrt{3}\), which is irrational.
Step 3
Exam Tip
The non-zero condition is important because multiplication by (0) gives (0). चरण 1: अशून्य परिमेय संख्या से अपरिमेय संख्या को गुणा करने पर परिणाम अपरिमेय रहता है। चरण 2: जैसे \(2 \times \sqrt{3}=2\sqrt{3}\), जो अपरिमेय है। चरण 3: यहां अशून्य शर्त जरूरी है, क्योंकि शून्य से गुणा करने पर परिणाम (0) होगा।
(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।
\(\sqrt{2}\times\sqrt{5}=\sqrt{10}\), which is irrational. Multiplying equal roots can often give a rational number.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\times\sqrt{5}\). \(\sqrt{2}\times\sqrt{5}=\sqrt{10}\), which is irrational. Multiplying equal roots can often give a rational number.
Step 3
Exam Tip
\(\sqrt{2}\times\sqrt{5}=\sqrt{10}\) है जो अपरिमेय है। समान जड़ों का गुणन अक्सर परिमेय दे सकता है।
A. जब (a) कोई भी परिमेय संख्या हो/When (a) is any rational number
Step 1
Concept
Adding a rational number to an irrational number gives an irrational result. This simple property often appears in MCQs.
Step 2
Why this answer is correct
The correct answer is A. जब (a) कोई भी परिमेय संख्या हो / When (a) is any rational number. Adding a rational number to an irrational number gives an irrational result. This simple property often appears in MCQs.
Step 3
Exam Tip
परिमेय में अपरिमेय जोड़ने पर परिणाम अपरिमेय रहता है। यह आसान गुण अक्सर MCQ में आता है।
\(\sqrt{5}\) is irrational and \(2-\sqrt{5}\) is also irrational.
Step 2
Why this answer is correct
Their sum is (2) which is rational.
Step 3
Exam Tip
There is no single always rule for the sum of two irrational numbers. चरण 1: \(\sqrt{5}\) अपरिमेय है और \(2-\sqrt{5}\) भी अपरिमेय है। चरण 2: उनका योग (2) है जो परिमेय है। चरण 3: दो अपरिमेय संख्याओं के योग के लिए एक ही नियम हर बार लागू नहीं होता।
Its reciprocal \(\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}\) is also irrational.
Step 3
Exam Tip
Do not assume the reciprocal of a non-zero irrational surd is rational. चरण 1: \(\sqrt{12}=2\sqrt{3}\) अपरिमेय है। चरण 2: इसका व्युत्क्रम \(\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}\) भी अपरिमेय है। चरण 3: अशून्य अपरिमेय मूल के व्युत्क्रम को परिमेय मानने की गलती न करें।
The sum of two irrational numbers can be rational.
Step 2
Why this answer is correct
For example, (\sqrt{2}+\(-\sqrt{2}\)=0). Therefore, saying (a+b) is always irrational is false.
Step 3
Exam Tip
Be careful with universal statements about two irrational numbers. चरण 1: दो अपरिमेय संख्याओं का योग कभी परिमेय भी हो सकता है। चरण 2: उदाहरण (\sqrt{2}+\(-\sqrt{2}\)=0) है। इसलिए (a+b) हमेशा अपरिमेय कहना गलत है। चरण 3: दो अपरिमेय संख्याओं पर हमेशा वाले नियम बहुत सावधानी से लगाएँ।
A rational number minus an irrational number is irrational.
Step 2
Why this answer is correct
If (r-s) were rational, then (s=r-(r-s)) would be rational, which is impossible.
Step 3
Exam Tip
Use the same reasoning for subtraction as for addition. चरण 1: परिमेय संख्या में से अपरिमेय संख्या घटाने पर परिणाम अपरिमेय रहता है। चरण 2: यदि (r-s) परिमेय हो, तो (s=r-(r-s)) परिमेय हो जाएगा, जो असंभव है। चरण 3: घटाव में भी वही सोच रखें जो योग में रखते हैं।
Substituting (x=0) in \(4x^3-7x\) gives (0), and it is not the zero polynomial. For (x=0), the constant term must be (0).
Step 2
Why this answer is correct
The correct answer is B. \(4x^3-7x\). Substituting (x=0) in \(4x^3-7x\) gives (0), and it is not the zero polynomial. For (x=0), the constant term must be (0).
Step 3
Exam Tip
\(4x^3-7x\) में (x=0) रखने पर (0) मिलता है और यह शून्य बहुपद नहीं है। (x=0) के लिए अचर पद (0) होना चाहिए।
With rational coefficients \(a+\sqrt{b}\) is accompanied by \(a-\sqrt{b}\). In exams identify conjugate zeroes quickly.
Step 2
Why this answer is correct
The correct answer is A. \(4-\sqrt{11}\). With rational coefficients \(a+\sqrt{b}\) is accompanied by \(a-\sqrt{b}\). In exams identify conjugate zeroes quickly.
Step 3
Exam Tip
परिमेय गुणांकों में \(a+\sqrt{b}\) के साथ \(a-\sqrt{b}\) भी शून्यक होता है। परीक्षा में संयुग्मी शून्यक तुरंत पहचानें।
The companion zero is \(5-2\sqrt{6}\), with sum (10) and product (25-24=1). In exams form the polynomial using the conjugate.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-10x+1\). The companion zero is \(5-2\sqrt{6}\), with sum (10) and product (25-24=1). In exams form the polynomial using the conjugate.
Step 3
Exam Tip
साथी शून्यक \(5-2\sqrt{6}\) होगा, योग (10) और गुणनफल (25-24=1) है। परीक्षा में संयुग्मी लेकर बहुपद बनाएं।
With rational coefficients, \(a+\sqrt{b}\) is accompanied by \(a-\sqrt{b}\). In exams identify conjugate zeroes quickly.
Step 2
Why this answer is correct
The correct answer is A. \(2-\sqrt{7}\). With rational coefficients, \(a+\sqrt{b}\) is accompanied by \(a-\sqrt{b}\). In exams identify conjugate zeroes quickly.
Step 3
Exam Tip
परिमेय गुणांकों में \(a+\sqrt{b}\) के साथ \(a-\sqrt{b}\) भी शून्यक आता है। परीक्षा में संयुग्मी शून्यकों को तुरंत पहचानें।
For rational coefficients, irrational zeroes usually occur in conjugate pairs. Hence the companion zero of \(3-\sqrt{5}\) is \(3+\sqrt{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(3+\sqrt{5}\). For rational coefficients, irrational zeroes usually occur in conjugate pairs. Hence the companion zero of \(3-\sqrt{5}\) is \(3+\sqrt{5}\).
Step 3
Exam Tip
परिमेय गुणांकों में अपरिमेय शून्यक सामान्यतः संयुग्मी रूप में आते हैं। इसलिए \(3-\sqrt{5}\) का साथी शून्यक \(3+\sqrt{5}\) होगा।
(p(x)=\(x+\sqrt{5}\)2), so the zero is \(-\sqrt{5}\) twice. A perfect-square form gives a repeated zero.
Step 2
Why this answer is correct
The correct answer is A. \(-\sqrt{5}\) दो बार / \(-\sqrt{5}\) twice. (p(x)=\(x+\sqrt{5}\)2), so the zero is \(-\sqrt{5}\) twice. A perfect-square form gives a repeated zero.
Step 3
Exam Tip
(p(x)=\(x+\sqrt{5}\)2), इसलिए शून्यक \(-\sqrt{5}\) दो बार है। पूर्ण वर्ग रूप से दोहराया शून्यक मिलता है।
With rational coefficients, the conjugate of an irrational zero is also a zero. So \(2-\sqrt{3}\) will be the other zero.
Step 2
Why this answer is correct
The correct answer is A. \(2-\sqrt{3}\). With rational coefficients, the conjugate of an irrational zero is also a zero. So \(2-\sqrt{3}\) will be the other zero.
Step 3
Exam Tip
परिमेय गुणांकों में अपरिमेय शून्यक का संयुग्मी भी शून्यक होता है। इसलिए \(2-\sqrt{3}\) दूसरा शून्यक होगा।
Both zeroes are \(\sqrt{3}\), so the sum is \(2\sqrt{3}\). In \(x^2+kx+3\), the sum is (-k), hence \(k=-2\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(-2\sqrt{3}\). Both zeroes are \(\sqrt{3}\), so the sum is \(2\sqrt{3}\). In \(x^2+kx+3\), the sum is (-k), hence \(k=-2\sqrt{3}\).
Step 3
Exam Tip
दोनों शून्यक \(\sqrt{3}\) हैं, इसलिए योग \(2\sqrt{3}\) है। \(x^2+kx+3\) में योग (-k) है, अतः \(k=-2\sqrt{3}\)।
For a quadratic with rational coefficients, \(a-\sqrt{b}\) accompanies \(a+\sqrt{b}\). Remember this as the conjugate-zero rule.
Step 2
Why this answer is correct
The correct answer is A. \(3-\sqrt{5}\). For a quadratic with rational coefficients, \(a-\sqrt{b}\) accompanies \(a+\sqrt{b}\). Remember this as the conjugate-zero rule.
Step 3
Exam Tip
परिमेय गुणांकों वाले द्विघात में \(a+\sqrt{b}\) के साथ \(a-\sqrt{b}\) भी शून्यक होता है। परीक्षा में इसे संयुग्मी शून्यक नियम की तरह याद रखें।
\(\frac{\sqrt{3}}{3}\) is irrational and lies between (0) and (1). Dividing by a non zero rational keeps irrationality.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\sqrt{3}}{3}\). \(\frac{\sqrt{3}}{3}\) is irrational and lies between (0) and (1). Dividing by a non zero rational keeps irrationality.
Step 3
Exam Tip
\(\frac{\sqrt{3}}{3}\) अपरिमेय है और इसका मान (0) और (1) के बीच है। गैर शून्य परिमेय से भाग देने पर अपरिमेयता रहती है।
\(\frac{\sqrt{2}}{2}\) is irrational and its value lies between (0) and (1). The root part makes it irrational.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\sqrt{2}}{2}\). \(\frac{\sqrt{2}}{2}\) is irrational and its value lies between (0) and (1). The root part makes it irrational.
Step 3
Exam Tip
\(\frac{\sqrt{2}}{2}\) अपरिमेय है और इसका मान (0) और (1) के बीच है। जड़ वाला भाग इसे अपरिमेय बनाता है।
\(\sqrt{2}\) is irrational, and dividing by non-zero rational (2) keeps it irrational.
Step 2
Why this answer is correct
\(\frac{\sqrt{2}}{2}\) is about (0.707), so it lies between (0) and (1).
Step 3
Exam Tip
Check both the interval condition and the nature of the number. चरण 1: \(\sqrt{2}\) अपरिमेय है और अशून्य परिमेय (2) से भाग देने पर अपरिमेयता बनी रहती है। चरण 2: \(\frac{\sqrt{2}}{2}\) लगभग (0.707) है, इसलिए यह (0) और (1) के बीच है। चरण 3: अंतराल और संख्या की प्रकृति दोनों शर्तें साथ-साथ जाँचें।
(x+0=x), so the nature remains the same and it is irrational.
Step 3
Exam Tip
Adding zero does not change either the value or the type of a number. चरण 1: (0) परिमेय संख्या है। चरण 2: (x+0=x), इसलिए संख्या की प्रकृति वही रहेगी और वह अपरिमेय होगी। चरण 3: शून्य जोड़ने से मान और प्रकृति दोनों नहीं बदलते।
\(0\times\sqrt{7}=0\), \(\sqrt{7}\times\sqrt{7}=7\), and \(\sqrt{28}\div\sqrt{7}=2\) are rational.
Step 2
Why this answer is correct
\(4\sqrt{7}\) is a non-zero rational multiple of an irrational number, so it is irrational.
Step 3
Exam Tip
Quickly identify multiplication by zero as rational. चरण 1: \(0\times\sqrt{7}=0\), \(\sqrt{7}\times\sqrt{7}=7\), और \(\sqrt{28}\div\sqrt{7}=2\) परिमेय हैं। चरण 2: \(4\sqrt{7}\) में अशून्य परिमेय गुणक और अपरिमेय मूल है, इसलिए यह अपरिमेय है। चरण 3: शून्य से गुणा वाले विकल्प को जल्दी परिमेय पहचानें।
A. \(\sqrt{8}\) और \(-\sqrt{8}\)/\(\sqrt{8}\) and \(-\sqrt{8}\)
Step 1
Concept
(\sqrt{8}+\(-\sqrt{8}\)=0), which is rational. In exams one counterexample is enough to disprove a universal statement.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{8}\) और \(-\sqrt{8}\) / \(\sqrt{8}\) and \(-\sqrt{8}\). (\sqrt{8}+\(-\sqrt{8}\)=0), which is rational. In exams one counterexample is enough to disprove a universal statement.
Step 3
Exam Tip
(\sqrt{8}+\(-\sqrt{8}\)=0), जो परिमेय है। परीक्षा में गलत सार्वत्रिक कथन तोड़ने के लिए एक प्रतिउदाहरण काफी है।
(5) is a non zero rational number so \(5\sqrt{3}\) is irrational. Remember multiplication by (0) gives (0).
Step 2
Why this answer is correct
The correct answer is A. \(5\times\sqrt{3}\). (5) is a non zero rational number so \(5\sqrt{3}\) is irrational. Remember multiplication by (0) gives (0).
Step 3
Exam Tip
(5) गैर शून्य परिमेय है इसलिए \(5\sqrt{3}\) अपरिमेय है। ध्यान रखें (0) से गुणा करने पर परिणाम (0) होता है।
A. \(\sqrt{12}\) और \(\sqrt{3}\)/\(\sqrt{12}\) and \(\sqrt{3}\)
Step 1
Concept
\(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{3}\) are both irrational.
Step 2
Why this answer is correct
Their product is \(\sqrt{36}=6\), which is rational, and their sum is \(3\sqrt{3}\), which is irrational.
Step 3
Exam Tip
Check the nature of the sum and product separately. चरण 1: \(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{3}\) दोनों अपरिमेय हैं। चरण 2: उनका गुणन \(\sqrt{36}=6\) परिमेय है, और योग \(3\sqrt{3}\) अपरिमेय है। चरण 3: योग और गुणन की प्रकृति अलग-अलग जाँचें।
B. \(\sqrt{3}\) और \(2\sqrt{3}\)/\(\sqrt{3}\) and \(2\sqrt{3}\)
Step 1
Concept
\(\sqrt{3}\) and \(2\sqrt{3}\) are both irrational.
Step 2
Why this answer is correct
Their sum is \(3\sqrt{3}\), which is irrational.
Step 3
Exam Tip
In sum questions, identify whether like surds cancel or combine. चरण 1: \(\sqrt{3}\) और \(2\sqrt{3}\) दोनों अपरिमेय हैं। चरण 2: उनका योग \(3\sqrt{3}\) है, जो अपरिमेय है। चरण 3: योग वाले प्रश्नों में कटने वाले और जुड़ने वाले समान मूल अलग-अलग पहचानें।
(6) is a non-zero rational number and \(\sqrt{19}\) is irrational.
Step 2
Why this answer is correct
\(6\sqrt{19}\) remains irrational.
Step 3
Exam Tip
Multiplication by zero is a special case, so focus on non-zero rational factors. चरण 1: (6) अशून्य परिमेय है और \(\sqrt{19}\) अपरिमेय है। चरण 2: \(6\sqrt{19}\) अपरिमेय रहेगा। चरण 3: शून्य से गुणा करने का मामला अलग है, इसलिए अशून्य परिमेय पर ध्यान दें।
(4) is a non-zero rational number and \(\sqrt{13}\) is irrational.
Step 2
Why this answer is correct
\(4\sqrt{13}\) remains irrational.
Step 3
Exam Tip
Multiplication by zero is a special case, so focus on non-zero rational factors. चरण 1: (4) अशून्य परिमेय है और \(\sqrt{13}\) अपरिमेय है। चरण 2: \(4\sqrt{13}\) अपरिमेय रहेगा। चरण 3: शून्य से गुणा करने का मामला अलग है, इसलिए अशून्य परिमेय पर ध्यान दें।
(3) is a non-zero rational number and \(\sqrt{7}\) is irrational.
Step 2
Why this answer is correct
\(3\sqrt{7}\) remains irrational.
Step 3
Exam Tip
Multiplication by zero is a special case, so focus on non-zero rational factors. चरण 1: (3) अशून्य परिमेय है और \(\sqrt{7}\) अपरिमेय है। चरण 2: \(3\sqrt{7}\) अपरिमेय रहेगा। चरण 3: शून्य से गुणा करने का मामला अलग होता है, इसलिए अशून्य परिमेय पर ध्यान दें।
A. दूसरा (7), कटान ((6,0)), ((7,0))/Other (7), intersections ((6,0)), ((7,0))
Step 1
Concept
In the quadratic, the sum of zeroes is (13), so the other zero is (7). Tip: convert a zero into ((x,0)).
Step 2
Why this answer is correct
The correct answer is A. दूसरा (7), कटान ((6,0)), ((7,0)) / Other (7), intersections ((6,0)), ((7,0)). In the quadratic, the sum of zeroes is (13), so the other zero is (7). Tip: convert a zero into ((x,0)).
Step 3
Exam Tip
द्विघात में शून्यकों का योग (13) है, इसलिए दूसरा शून्यक (7) है। टिप: शून्यक को ((x,0)) में बदलें।
The average of the two zeroes is (5), so the other zero is (11). Tip: the axis of symmetry passes through the midpoint of zeroes.
Step 2
Why this answer is correct
The correct answer is A. (11). The average of the two zeroes is (5), so the other zero is (11). Tip: the axis of symmetry passes through the midpoint of zeroes.
Step 3
Exam Tip
दो शून्यकों का औसत (5) है इसलिए दूसरा शून्यक (11) होगा। टिप: सममिति अक्ष शून्यकों के मध्य से गुजरता है।
A. दूसरा (7), कटान ((4,0)), ((7,0))/Other (7), intersections ((4,0)), ((7,0))
Step 1
Concept
In the quadratic, the sum of zeroes is (11), so the other zero is (7). Tip: convert a zero into ((x,0)).
Step 2
Why this answer is correct
The correct answer is A. दूसरा (7), कटान ((4,0)), ((7,0)) / Other (7), intersections ((4,0)), ((7,0)). In the quadratic, the sum of zeroes is (11), so the other zero is (7). Tip: convert a zero into ((x,0)).
Step 3
Exam Tip
द्विघात में शून्यकों का योग (11) है, इसलिए दूसरा शून्यक (7) है। टिप: शून्यक को ((x,0)) में बदलें।
The average of the two zeroes is (-2), so the other zero is (-9). Tip: connect the axis of symmetry with the midpoint of zeroes.
Step 2
Why this answer is correct
The correct answer is A. (-9). The average of the two zeroes is (-2), so the other zero is (-9). Tip: connect the axis of symmetry with the midpoint of zeroes.
Step 3
Exam Tip
दो शून्यकों का औसत (-2) है, इसलिए दूसरा शून्यक (-9) होगा। टिप: सममिति अक्ष को शून्यकों के मध्य से जोड़ें।
A. दूसरा (5), कटान ((4,0)), ((5,0))/Other (5), intersections ((4,0)), ((5,0))
Step 1
Concept
In the quadratic, the sum of zeroes is (9), so the other zero is (5). Tip: quickly convert a zero to ((x,0)).
Step 2
Why this answer is correct
The correct answer is A. दूसरा (5), कटान ((4,0)), ((5,0)) / Other (5), intersections ((4,0)), ((5,0)). In the quadratic, the sum of zeroes is (9), so the other zero is (5). Tip: quickly convert a zero to ((x,0)).
Step 3
Exam Tip
द्विघात में शून्यकों का योग (9) है इसलिए दूसरा शून्यक (5) है। टिप: शून्यक को तुरंत ((x,0)) में बदलें।
The average of the two zeroes is (3), so the other zero is (-5). Tip: set \(\frac{a+b}{2}\) equal to the axis of symmetry.
Step 2
Why this answer is correct
The correct answer is A. (-5). The average of the two zeroes is (3), so the other zero is (-5). Tip: set \(\frac{a+b}{2}\) equal to the axis of symmetry.
Step 3
Exam Tip
दो शून्यकों का औसत (3) है इसलिए दूसरा शून्यक (-5) होगा। टिप: \(\frac{a+b}{2}\) को सममिति अक्ष के बराबर रखें।
The average of the two zeroes is (4), so the other zero is (10). Tip: connect the axis of symmetry with the midpoint of zeroes.
Step 2
Why this answer is correct
The correct answer is A. (10). The average of the two zeroes is (4), so the other zero is (10). Tip: connect the axis of symmetry with the midpoint of zeroes.
Step 3
Exam Tip
दोनों शून्यकों का औसत (4) है इसलिए दूसरा शून्यक (10) होगा। टिप: सममिति अक्ष को शून्यकों के मध्य मान से जोड़ें।
A. दूसरा (4), कटान ((3,0)), ((4,0))/Other (4), intersections ((3,0)), ((4,0))
Step 1
Concept
In the quadratic, the sum of zeroes is (7), so the other zero is (4). Tip: quickly convert a zero to ((x,0)).
Step 2
Why this answer is correct
The correct answer is A. दूसरा (4), कटान ((3,0)), ((4,0)) / Other (4), intersections ((3,0)), ((4,0)). In the quadratic, the sum of zeroes is (7), so the other zero is (4). Tip: quickly convert a zero to ((x,0)).
Step 3
Exam Tip
द्विघात में शून्यकों का योग (7) है, इसलिए दूसरा शून्यक (4) है। टिप: शून्यक को तुरंत ((x,0)) में बदलें।
The average of the two zeroes is (2), so the other zero is (-5). Tip: set \( \frac{a+b}{2} \) equal to the axis of symmetry.
Step 2
Why this answer is correct
The correct answer is A. (-5). The average of the two zeroes is (2), so the other zero is (-5). Tip: set \( \frac{a+b}{2} \) equal to the axis of symmetry.
Step 3
Exam Tip
दो शून्यकों का औसत (2) है, इसलिए दूसरा शून्यक (-5) होगा। टिप: \( \frac{a+b}{2} \) को सममिति अक्ष के बराबर रखें।
The average of the two zeroes is (1), so the other zero is (7). Tip: the axis of symmetry passes through the midpoint of zeroes.
Step 2
Why this answer is correct
The correct answer is C. (7). The average of the two zeroes is (1), so the other zero is (7). Tip: the axis of symmetry passes through the midpoint of zeroes.
Step 3
Exam Tip
दो शून्यकों का औसत (1) होगा इसलिए दूसरा शून्यक (7) है। टिप: सममिति अक्ष शून्यकों के मध्य से गुजरता है।
A. दूसरा (3), कटान ((2,0)), ((3,0))/Other (3), intersections ((2,0)), ((3,0))
Step 1
Concept
In the quadratic, the sum of zeroes is (5), so the other zero is (3). Tip: immediately convert a zero to ((x,0)).
Step 2
Why this answer is correct
The correct answer is A. दूसरा (3), कटान ((2,0)), ((3,0)) / Other (3), intersections ((2,0)), ((3,0)). In the quadratic, the sum of zeroes is (5), so the other zero is (3). Tip: immediately convert a zero to ((x,0)).
Step 3
Exam Tip
द्विघात में शून्यकों का योग (5) है, इसलिए दूसरा शून्यक (3) है। टिप: शून्यक को तुरंत ((x,0)) में बदलें।
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।
The other zero will be \(6-\sqrt{19}\), and the product is (36-19=17). In exams connect the constant term with the product of zeroes.
Step 2
Why this answer is correct
The correct answer is A. (17). The other zero will be \(6-\sqrt{19}\), and the product is (36-19=17). In exams connect the constant term with the product of zeroes.
Step 3
Exam Tip
दूसरा शून्यक \(6-\sqrt{19}\) होगा और गुणनफल (36-19=17) है। परीक्षा में स्थिर पद को शून्यकों के गुणनफल से जोड़ें।
A. \(\sqrt{3}=-\frac{p}{q}\) होगा जो असंभव है/\(\sqrt{3}=-\frac{p}{q}\) would be true which is impossible
Step 1
Concept
\(-\frac{p}{q}\) is rational so it would make \(\sqrt{3}\) rational which is false. In exams recognize the contradiction method.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{3}=-\frac{p}{q}\) होगा जो असंभव है / \(\sqrt{3}=-\frac{p}{q}\) would be true which is impossible. \(-\frac{p}{q}\) is rational so it would make \(\sqrt{3}\) rational which is false. In exams recognize the contradiction method.
Step 3
Exam Tip
\(-\frac{p}{q}\) परिमेय है इसलिए इससे \(\sqrt{3}\) परिमेय हो जाएगा जो गलत है। परीक्षा में विरोधाभास विधि पहचानें।
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।
A. दोनों शून्यक \(\sqrt{5}\) हैं/Both zeroes are \(\sqrt{5}\)
Step 1
Concept
This polynomial equals (\(x-\sqrt{5}\)2). Hence \(\sqrt{5}\) is a repeated zero.
Step 2
Why this answer is correct
The correct answer is A. दोनों शून्यक \(\sqrt{5}\) हैं / Both zeroes are \(\sqrt{5}\). This polynomial equals (\(x-\sqrt{5}\)2). Hence \(\sqrt{5}\) is a repeated zero.
Step 3
Exam Tip
यह बहुपद (\(x-\sqrt{5}\)2) के बराबर है। अतः \(\sqrt{5}\) दोहरा शून्यक है।
(\(2+\sqrt{3}\)\(2-\sqrt{3}\)=4-3=1) which is rational. In exams remember conjugate multiplication as a counterexample.
Step 2
Why this answer is correct
The correct answer is A. (\(2+\sqrt{3}\)\(2-\sqrt{3}\)). (\(2+\sqrt{3}\)\(2-\sqrt{3}\)=4-3=1) which is rational. In exams remember conjugate multiplication as a counterexample.
Step 3
Exam Tip
(\(2+\sqrt{3}\)\(2-\sqrt{3}\)=4-3=1) है जो परिमेय है। परीक्षा में संयुग्मी गुणन को प्रतिउदाहरण के रूप में याद रखें।
A. \(\sqrt{2}\) और \(3\sqrt{2}\)/\(\sqrt{2}\) and \(3\sqrt{2}\)
Step 1
Concept
\(\sqrt{2}\cdot3\sqrt{2}=6\), which is rational. In exams remember counterexamples for products of irrational numbers.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\) और \(3\sqrt{2}\) / \(\sqrt{2}\) and \(3\sqrt{2}\). \(\sqrt{2}\cdot3\sqrt{2}=6\), which is rational. In exams remember counterexamples for products of irrational numbers.
Step 3
Exam Tip
\(\sqrt{2}\cdot3\sqrt{2}=6\), जो परिमेय है। परीक्षा में अपरिमेय संख्याओं के गुणनफल के लिए प्रतिउदाहरण याद रखें।
The like (x) terms cancel and the value left is \(2\sqrt{2}\). In exams do not be confused by the type of number during algebraic simplification.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{2}\). The like (x) terms cancel and the value left is \(2\sqrt{2}\). In exams do not be confused by the type of number during algebraic simplification.
Step 3
Exam Tip
समान (x) पद कट जाते हैं और मान \(2\sqrt{2}\) बचता है। परीक्षा में बीजीय सरलीकरण में संख्या के प्रकार से भ्रमित न हों।
A. \(2+\sqrt{5}\) और \(2-\sqrt{5}\)/\(2+\sqrt{5}\) and \(2-\sqrt{5}\)
Step 1
Concept
The sum is (\(2+\sqrt{5}\)+\(2-\sqrt{5}\)=4), which is rational. In exams remember conjugate pairs as counterexamples.
Step 2
Why this answer is correct
The correct answer is A. \(2+\sqrt{5}\) और \(2-\sqrt{5}\) / \(2+\sqrt{5}\) and \(2-\sqrt{5}\). The sum is (\(2+\sqrt{5}\)+\(2-\sqrt{5}\)=4), which is rational. In exams remember conjugate pairs as counterexamples.
Step 3
Exam Tip
योग (\(2+\sqrt{5}\)+\(2-\sqrt{5}\)=4) है, जो परिमेय है। परीक्षा में संयुग्मी जोड़ों को प्रतिउदाहरण के रूप में याद रखें।
A. (m) पूर्ण वर्ग नहीं है/(m) is not a perfect square
Step 1
Concept
The square root of a perfect square is an integer, so for an irrational square root (m) is not a perfect square. In exams identifying perfect squares is important.
Step 2
Why this answer is correct
The correct answer is A. (m) पूर्ण वर्ग नहीं है / (m) is not a perfect square. The square root of a perfect square is an integer, so for an irrational square root (m) is not a perfect square. In exams identifying perfect squares is important.
Step 3
Exam Tip
पूर्ण वर्ग का वर्गमूल पूर्णांक होता है, इसलिए अपरिमेय वर्गमूल के लिए (m) पूर्ण वर्ग नहीं होगा। परीक्षा में पूर्ण वर्ग पहचानना जरूरी 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।
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।
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।
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।
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।
(180) is not a perfect square so \(\sqrt{180}\) is irrational. Subtracting an irrational from a rational gives an irrational result.
Step 2
Why this answer is correct
The correct answer is A. (m=180). (180) is not a perfect square so \(\sqrt{180}\) is irrational. Subtracting an irrational from a rational gives an irrational result.
Step 3
Exam Tip
(180) पूर्ण वर्ग नहीं है इसलिए \(\sqrt{180}\) अपरिमेय है। परिमेय से अपरिमेय घटाने पर परिणाम अपरिमेय होता है।
\(\sqrt{8}=2\sqrt{2}\), so the sum is \(3\sqrt{2}\). A non zero rational multiple of \(\sqrt{2}\) remains irrational.
Step 2
Why this answer is correct
The correct answer is A. यह \(3\sqrt{2}\) है / It is \(3\sqrt{2}\). \(\sqrt{8}=2\sqrt{2}\), so the sum is \(3\sqrt{2}\). A non zero rational multiple of \(\sqrt{2}\) remains irrational.
Step 3
Exam Tip
\(\sqrt{8}=2\sqrt{2}\), इसलिए योग \(3\sqrt{2}\) है। गैर शून्य परिमेय गुणक के साथ \(\sqrt{2}\) अपरिमेय रहता है।
A. (k) पूर्ण वर्ग नहीं है/(k) is not a perfect square
Step 1
Concept
If a positive integer is not a perfect square its square root is irrational. So (k) is not a perfect square.
Step 2
Why this answer is correct
The correct answer is A. (k) पूर्ण वर्ग नहीं है / (k) is not a perfect square. If a positive integer is not a perfect square its square root is irrational. So (k) is not a perfect square.
Step 3
Exam Tip
धनात्मक पूर्णांक पूर्ण वर्ग न हो तो उसकी वर्गमूल अपरिमेय होती है। इसलिए (k) पूर्ण वर्ग नहीं होगा।
