100 results found for "square-root-interval" in Class 10.
यदि संख्या रेखा पर \(x=-\sqrt{29}\), तो (x) किस अंतराल में होगा?
If \(x=-\sqrt{29}\) on the number line, in which interval will (x) lie?
#number-line
#negative-square-root
#interval
A ( -5 ) और ( -4 ) के बीच / Between ( -5 ) and ( -4 )
B ( -6 ) और ( -5 ) के बीच / Between ( -6 ) and ( -5 )
C (4) और (5) के बीच / Between (4) and (5)
D (5) और (6) के बीच / Between (5) and (6)
Explanation opens after your attempt
Correct Answer
B. ( -6 ) और ( -5 ) के बीच / Between ( -6 ) and ( -5 )
Step 1
Concept
Since \(5<\sqrt{29}<6\), \(-6<-\sqrt{29}<-5\). For negative roots, write the reversed interval carefully.
Step 2
Why this answer is correct
The correct answer is B. ( -6 ) और ( -5 ) के बीच / Between ( -6 ) and ( -5 ). Since \(5<\sqrt{29}<6\), \(-6<-\sqrt{29}<-5\). For negative roots, write the reversed interval carefully.
Step 3
Exam Tip
क्योंकि \(5<\sqrt{29}<6\), इसलिए \(-6<-\sqrt{29}<-5\)। ऋणात्मक मूलों में क्रम उलटकर लिखें।
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किस विकल्प में \(\sqrt{20}\) का संख्या रेखा पर सही अंतराल दिया गया है?
Which option gives the correct interval for \(\sqrt{20}\) on the number line?
#polynomials
#number-line
#square-root
#interval
A ((4,5))
B ((3,4))
C ((5,6))
D ((2,3))
Explanation opens after your attempt
Correct Answer
A. ((4,5))
Step 1
Concept
Because \(4^2=16\) and \(5^2=25\), \(4<\sqrt{20}<5\). Perfect-square bounds quickly give the interval.
Step 2
Why this answer is correct
The correct answer is A. ((4,5)). Because \(4^2=16\) and \(5^2=25\), \(4<\sqrt{20}<5\). Perfect-square bounds quickly give the interval.
Step 3
Exam Tip
क्योंकि \(4^2=16\) और \(5^2=25\), इसलिए \(4<\sqrt{20}<5\)। पूर्ण वर्गों की सीमा तुरंत अंतराल देती है।
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किस समीकरण में (x=0) एक मूल है और दूसरा मूल ऋणात्मक है?
In which equation is (x=0) one root and the other root negative?
#quadratic-equations
#zero-root
#root-sign
#hard
A \(x^2+7x=0\)
B \(x^2-7x=0\)
C \(x^2+7=0\)
D \(x^2-7=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+7x=0\)
Step 1
Concept
(x-2 +7x=x(x+7)), so the roots are (0) and (-7). The other root is negative.
Step 2
Why this answer is correct
The correct answer is A. \(x^2+7x=0\). (x-2 +7x=x(x+7)), so the roots are (0) and (-7). The other root is negative.
Step 3
Exam Tip
(x-2 +7x=x(x+7)), इसलिए मूल (0) और (-7) हैं। दूसरा मूल ऋणात्मक है।
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किस समीकरण में (x=0) एक मूल है और दूसरा मूल धनात्मक है?
In which equation is (x=0) one root and the other root positive?
#quadratic-equations
#zero-root
#root-sign
#hard
A \(x^2-6x=0\)
B \(x^2+6x=0\)
C \(x^2+6=0\)
D \(x^2-6=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2-6x=0\)
Step 1
Concept
(x-2 -6x=x(x-6)), so the roots are (0) and (6). The other root is positive.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-6x=0\). (x-2 -6x=x(x-6)), so the roots are (0) and (6). The other root is positive.
Step 3
Exam Tip
(x-2 -6x=x(x-6)), इसलिए मूल (0) और (6) हैं। दूसरा मूल धनात्मक है।
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कौन सा वर्गमूल परिमेय है, इसलिए उस पर \(\sqrt{2}\) जैसी अपरिमेयता सिद्धि लागू नहीं होती?
Which square root is rational, so an irrationality proof like \(\sqrt{2}\) does not apply to it?
#perfect square
#rational square root
#class 10
A \(\sqrt{2}\)
B \(\sqrt{3}\)
C \(\sqrt{4}\)
D \(\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{4}\)
Step 1
Concept
(4) is a perfect square.
Step 2
Why this answer is correct
\(\sqrt{4}=2\), which is rational.
Step 3
Exam Tip
Square roots of perfect squares are rational. चरण 1: (4) पूर्ण वर्ग है। चरण 2: \(\sqrt{4}=2\), जो परिमेय संख्या है। चरण 3: पूर्ण वर्गों के वर्गमूल परिमेय होते हैं।
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कौन सा वर्गमूल इस अध्याय की अपरिमेयता सिद्धि का उदाहरण नहीं है क्योंकि वह परिमेय है?
Which square root is not an example of irrationality proof in this chapter because it is rational?
#rational square root
#perfect square
#class 10
A \(\sqrt{9}\)
B \(\sqrt{2}\)
C \(\sqrt{3}\)
D \(\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{9}\)
Step 1
Concept
(9) is a perfect square.
Step 2
Why this answer is correct
\(\sqrt{9}=3\), which is rational.
Step 3
Exam Tip
A square root of a perfect square does not need an irrationality proof. चरण 1: (9) पूर्ण वर्ग है। चरण 2: \(\sqrt{9}=3\), जो परिमेय है। चरण 3: पूर्ण वर्ग के वर्गमूल को अपरिमेय सिद्ध करने की जरूरत नहीं होती।
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कौन सा वर्गमूल परिमेय है, इसलिए \(\sqrt{2}\), \(\sqrt{3}\), \(\sqrt{5}\) जैसी सिद्धि की जरूरत नहीं है?
Which square root is rational, so it does not need a proof like \(\sqrt{2}\), \(\sqrt{3}\), or \(\sqrt{5}\)?
#rational square root
#perfect square
#class 10
A \(\sqrt{4}\)
B \(\sqrt{2}\)
C \(\sqrt{3}\)
D \(\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{4}\)
Step 1
Concept
(4) is a perfect square.
Step 2
Why this answer is correct
\(\sqrt{4}=2\), which is rational.
Step 3
Exam Tip
The square root of a perfect square is not proved irrational. चरण 1: (4) पूर्ण वर्ग है। चरण 2: \(\sqrt{4}=2\), जो परिमेय संख्या है। चरण 3: पूर्ण वर्ग के वर्गमूल को अपरिमेय सिद्ध नहीं करते।
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निम्न में से कौन सी संख्या पूर्ण वर्ग नहीं है और उसका वर्गमूल अपरिमेय सिद्ध किया जाता है?
Which of the following is not a perfect square and its square root is proved irrational?
#perfect square
#sqrt2
#class 10
A (2)
B (4)
C (9)
D (25)
Explanation opens after your attempt
Step 1
Concept
(4), (9), and (25) are perfect squares.
Step 2
Why this answer is correct
(2) is not a perfect square, so \(\sqrt{2}\) is proved irrational.
Step 3
Exam Tip
First identify perfect and non-perfect squares. चरण 1: (4), (9) और (25) पूर्ण वर्ग हैं। चरण 2: (2) पूर्ण वर्ग नहीं है, इसलिए \(\sqrt{2}\) अपरिमेय सिद्ध किया जाता है। चरण 3: पूर्ण वर्ग और अपूर्ण वर्ग की पहचान पहले करें।
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संख्या रेखा पर \(6-\sqrt{39}\) का सही स्थान किस अंतराल में है?
In which interval is \(6-\sqrt{39}\) correctly located on the number line?
#number-line
#root-expression
#interval
A ( -1 ) और (0) के बीच / Between ( -1 ) and (0)
B (0) और (1) के बीच / Between (0) and (1)
C (1) और (2) के बीच / Between (1) and (2)
D ( -2 ) और ( -1 ) के बीच / Between ( -2 ) and ( -1 )
Explanation opens after your attempt
Correct Answer
A. ( -1 ) और (0) के बीच / Between ( -1 ) and (0)
Step 1
Concept
\( \sqrt{39}\approx6.245 \), so \(6-\sqrt{39}\approx-0.245\). Always check the sign in root subtraction.
Step 2
Why this answer is correct
The correct answer is A. ( -1 ) और (0) के बीच / Between ( -1 ) and (0). \( \sqrt{39}\approx6.245 \), so \(6-\sqrt{39}\approx-0.245\). Always check the sign in root subtraction.
Step 3
Exam Tip
\( \sqrt{39}\approx6.245 \), इसलिए \(6-\sqrt{39}\approx-0.245\) है। घटाव वाले मूल में चिह्न जरूर जाँचें।
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संख्या रेखा पर \(5-\sqrt{31}\) का सही स्थान किस अंतराल में है?
In which interval is \(5-\sqrt{31}\) correctly located on the number line?
#number-line
#root-expression
#interval
A (0) और (1) के बीच / Between (0) and (1)
B ( -1 ) और (0) के बीच / Between ( -1 ) and (0)
C (1) और (2) के बीच / Between (1) and (2)
D ( -2 ) और ( -1 ) के बीच / Between ( -2 ) and ( -1 )
Explanation opens after your attempt
Correct Answer
B. ( -1 ) और (0) के बीच / Between ( -1 ) and (0)
Step 1
Concept
\( \sqrt{31}\approx5.568 \), so \(5-\sqrt{31}\approx-0.568\). Always check the sign in root subtraction.
Step 2
Why this answer is correct
The correct answer is B. ( -1 ) और (0) के बीच / Between ( -1 ) and (0). \( \sqrt{31}\approx5.568 \), so \(5-\sqrt{31}\approx-0.568\). Always check the sign in root subtraction.
Step 3
Exam Tip
\( \sqrt{31}\approx5.568 \), इसलिए \(5-\sqrt{31}\approx-0.568\) है। घटाव वाले मूलों में चिह्न अवश्य जाँचें।
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यदि \(x=-5+\sqrt{22}\), तो संख्या रेखा पर (x) किस अंतराल में है?
If \(x=-5+\sqrt{22}\), in which interval is (x) on the number line?
#number-line
#root-expression
#interval
A ( -2 ) और ( -1 ) के बीच / Between ( -2 ) and ( -1 )
B (0) और (1) के बीच / Between (0) and (1)
C ( -1 ) और (0) के बीच / Between ( -1 ) and (0)
D (1) और (2) के बीच / Between (1) and (2)
Explanation opens after your attempt
Correct Answer
C. ( -1 ) और (0) के बीच / Between ( -1 ) and (0)
Step 1
Concept
Since \(4<\sqrt{22}<5\), \(-1<-5+\sqrt{22}<0\). Add bounds carefully in mixed expressions.
Step 2
Why this answer is correct
The correct answer is C. ( -1 ) और (0) के बीच / Between ( -1 ) and (0). Since \(4<\sqrt{22}<5\), \(-1<-5+\sqrt{22}<0\). Add bounds carefully in mixed expressions.
Step 3
Exam Tip
\(4<\sqrt{22}<5\), इसलिए \(-1<-5+\sqrt{22}<0\)। मिश्रित अभिव्यक्ति में सीमा जोड़ें।
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संख्या रेखा पर \(2-\sqrt{10}\) का सही स्थान किस अंतराल में है?
In which interval is \(2-\sqrt{10}\) correctly located on the number line?
#number-line
#root-expression
#interval
A (-2) और (-1) के बीच / Between (-2) and (-1)
B (-1) और (0) के बीच / Between (-1) and (0)
C (0) और (1) के बीच / Between (0) and (1)
D (1) और (2) के बीच / Between (1) and (2)
Explanation opens after your attempt
Correct Answer
A. (-2) और (-1) के बीच / Between (-2) and (-1)
Step 1
Concept
\( \sqrt{10}\approx3.162 \), so \(2-\sqrt{10}\approx-1.162\). Estimation is important in root subtraction.
Step 2
Why this answer is correct
The correct answer is A. (-2) और (-1) के बीच / Between (-2) and (-1). \( \sqrt{10}\approx3.162 \), so \(2-\sqrt{10}\approx-1.162\). Estimation is important in root subtraction.
Step 3
Exam Tip
\( \sqrt{10}\approx3.162 \) इसलिए \(2-\sqrt{10}\approx-1.162\) है। घटाव वाले मूल में अनुमान जरूरी है।
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संख्या रेखा पर (x) का स्थान ( -4 ) से \( \sqrt{13} \) इकाई दाईं ओर है। (x) किस अंतराल में है?
The point (x) is \( \sqrt{13} \) units to the right of (-4) on the number line. In which interval does (x) lie?
#number-line
#interval
#root-expression
A (-1) और (0) के बीच / Between (-1) and (0)
B (0) और (1) के बीच / Between (0) and (1)
C (-2) और (-1) के बीच / Between (-2) and (-1)
D (-4) और (-3) के बीच / Between (-4) and (-3)
Explanation opens after your attempt
Correct Answer
A. (-1) और (0) के बीच / Between (-1) and (0)
Step 1
Concept
\(x=-4+\sqrt{13}\), and \(3<\sqrt{13}<4\), so (-1<x<0). Add bounds in combined expressions.
Step 2
Why this answer is correct
The correct answer is A. (-1) और (0) के बीच / Between (-1) and (0). \(x=-4+\sqrt{13}\), and \(3<\sqrt{13}<4\), so (-1<x<0). Add bounds in combined expressions.
Step 3
Exam Tip
\(x=-4+\sqrt{13}\) और \(3<\sqrt{13}<4\), इसलिए (-1<x<0)। संयुक्त अभिव्यक्ति में सीमा जोड़ें।
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\(7x^2=175\) को वर्गमूल विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(7x^2=175\) by square root method?
#quadratic
#square-root-method
#solutions
A \(x=\pm5\)
B (x=5)
C (x=-5)
D \(x=\pm25\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm5\)
Step 1
Concept
First \(x^2=25\), so \(x=\pm5\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm5\). First \(x^2=25\), so \(x=\pm5\). In exams, write both signs while taking square root.
Step 3
Exam Tip
पहले \(x^2=25\) मिलता है, इसलिए \(x=\pm5\) है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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\(5x^2=80\) को वर्गमूल विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(5x^2=80\) by square root method?
#quadratic
#square-root-method
#solutions
A \(x=\pm4\)
B (x=4)
C (x=-4)
D \(x=\pm16\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm4\)
Step 1
Concept
First \(x^2=16\), so \(x=\pm4\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm4\). First \(x^2=16\), so \(x=\pm4\). In exams, write both signs while taking square root.
Step 3
Exam Tip
पहले \(x^2=16\) मिलता है, इसलिए \(x=\pm4\) है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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\(3x^2=12\) को वर्गमूल विधि से हल करने पर मूल क्या होंगे?
What roots are obtained by solving \(3x^2=12\) by square root method?
#quadratic
#square-root-method
#solutions
A \(x=\pm2\)
B (x=2)
C (x=-2)
D \(x=\pm4\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm2\)
Step 1
Concept
First \(x^2=4\), so \(x=\pm2\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm2\). First \(x^2=4\), so \(x=\pm2\). In exams, write both signs while taking square root.
Step 3
Exam Tip
पहले \(x^2=4\) मिलता है, इसलिए \(x=\pm2\) है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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\(x^2=169\) को वर्गमूल विधि से हल करने पर क्या मिलेगा?
Solving \(x^2=169\) by square root method gives what?
#quadratic
#square-root-method
#common-mistake
A \(x=\pm13\)
B (x=13)
C (x=-13)
D \(x=\pm169\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm13\)
Step 1
Concept
\(x=\pm\sqrt{169}=\pm13\). In exams, writing only (13) is an incomplete answer.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm13\). \(x=\pm\sqrt{169}=\pm13\). In exams, writing only (13) is an incomplete answer.
Step 3
Exam Tip
\(x=\pm\sqrt{169}=\pm13\) होता है। परीक्षा में केवल (13) लिखना अधूरा उत्तर है।
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वर्गमूल विधि से \(x^2=144\) के हल क्या हैं?
By square root method, what are the solutions of \(x^2=144\)?
#quadratic
#square-root-method
#solutions
A \(x=\pm12\)
B (x=12)
C (x=-12)
D \(x=\pm72\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm12\)
Step 1
Concept
\(x=\pm\sqrt{144}=\pm12\). In exams, do not forget \(\pm\) while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm12\). \(x=\pm\sqrt{144}=\pm12\). In exams, do not forget \(\pm\) while taking square root.
Step 3
Exam Tip
\(x=\pm\sqrt{144}=\pm12\) होता है। परीक्षा में वर्गमूल लेते समय \(\pm\) लगाना न भूलें।
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\(x^2=121\) को वर्गमूल विधि से हल करने पर क्या मिलेगा?
Solving \(x^2=121\) by square root method gives what?
#quadratic
#square-root-method
#common-mistake
A \(x=\pm11\)
B (x=11)
C (x=-11)
D \(x=\pm121\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm11\)
Step 1
Concept
\(x=\pm\sqrt{121}=\pm11\). In exams, writing only (11) is an incomplete answer.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm11\). \(x=\pm\sqrt{121}=\pm11\). In exams, writing only (11) is an incomplete answer.
Step 3
Exam Tip
\(x=\pm\sqrt{121}=\pm11\) होता है। परीक्षा में केवल (11) लिखना अधूरा उत्तर है।
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वर्गमूल विधि से \(x^2=64\) के हल क्या हैं?
By square root method, what are the solutions of \(x^2=64\)?
#quadratic
#square-root-method
#solutions
A \(x=\pm8\)
B (x=8)
C (x=-8)
D \(x=\pm32\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm8\)
Step 1
Concept
\(x=\pm\sqrt{64}=\pm8\). In exams, write both signs while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm8\). \(x=\pm\sqrt{64}=\pm8\). In exams, write both signs while taking square root.
Step 3
Exam Tip
\(x=\pm\sqrt{64}=\pm8\) होता है। परीक्षा में वर्गमूल लेते समय दोनों चिन्ह लिखें।
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किस समीकरण को वर्गमूल विधि से सीधे हल किया जा सकता है?
Which equation can be solved directly by square root method?
#quadratic
#square-root-method
#method-selection
A ((x-2)2 =9)
B \(x^2+5x+6=0\)
C \(x^2+2x+1=0\)
D \(2x^2+3x+1=0\)
Explanation opens after your attempt
Correct Answer
A. ((x-2)2 =9)
Step 1
Concept
\(In ((x-2)^2=9), square root can be taken directly. In exams, recognize the form ((\)expression\()^2=k).\)
Step 2
Why this answer is correct
\(The correct answer is A. ((x-2)^2=9). In ((x-2)^2=9), square root can be taken directly. In exams, recognize the form ((\)expression\()^2=k).\)
Step 3
Exam Tip
((x-2)2 =9) में सीधे वर्गमूल लिया जा सकता है। परीक्षा में ((expression\()^2=k) रूप को पहचानें\)।
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\(x^2=49\) को वर्गमूल विधि से हल करने पर क्या मिलेगा?
Solving \(x^2=49\) by square root method gives what?
#quadratic
#square-root-method
#common-mistake
A \(x=\pm7\)
B (x=7)
C (x=-7)
D \(x=\pm49\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm7\)
Step 1
Concept
\(x=\pm\sqrt{49}=\pm7\). In exams, writing only the positive root is a common mistake.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm7\). \(x=\pm\sqrt{49}=\pm7\). In exams, writing only the positive root is a common mistake.
Step 3
Exam Tip
\(x=\pm\sqrt{49}=\pm7\) होता है। परीक्षा में केवल धनात्मक मूल लिखना सामान्य गलती है।
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यदि \(x^2=16\), तो वर्गमूल विधि से (x) का मान क्या होगा?
If \(x^2=16\), what is the value of (x) by square root method?
#quadratic
#square-root-method
#roots
A \(x=\pm4\)
B (x=4)
C (x=-4)
D \(x=\pm8\)
Explanation opens after your attempt
Correct Answer
A. \(x=\pm4\)
Step 1
Concept
From \(x^2=16\), \(x=\pm\sqrt{16}=\pm4\). In exams, do not forget \(\pm\) while taking square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm4\). From \(x^2=16\), \(x=\pm\sqrt{16}=\pm4\). In exams, do not forget \(\pm\) while taking square root.
Step 3
Exam Tip
\(x^2=16\) से \(x=\pm\sqrt{16}=\pm4\) मिलता है। परीक्षा में वर्गमूल लेते समय \(\pm\) लगाना न भूलें।
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यदि \(a=7+4\sqrt{3}\), तो कौन-सा विकल्प (a) का वर्गमूल दर्शाता है?
If \(a=7+4\sqrt{3}\), which option represents a square root of (a)?
#surd square root
#identity
#class 10
A \(2+\sqrt{3}\)
B \(2-\sqrt{3}\)
C \(\sqrt{7}+2\)
D \(\sqrt{3}+1\)
Explanation opens after your attempt
Correct Answer
A. \(2+\sqrt{3}\)
Step 1
Concept
(\(2+\sqrt{3}\)2 =4+4\sqrt{3}+3).
Step 2
Why this answer is correct
This equals \(7+4\sqrt{3}\).
Step 3
Exam Tip
In such questions, identify the form \(m+n+2\sqrt{mn}\). चरण 1: (\(2+\sqrt{3}\)2 =4+4\sqrt{3}+3)। चरण 2: यह \(7+4\sqrt{3}\) के बराबर है। चरण 3: ऐसे प्रश्नों में \(m+n+2\sqrt{mn}\) का रूप पहचानें।
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\(2^6 \times 5^2\) का वर्गमूल क्या होगा?
What is the square root of \(2^6 \times 5^2\)?
#real-numbers
#square-root
#prime-factorisation
A \(2^2 \times 5\)
B \(2^3 \times 5\)
C \(2^3 \times 5^2\)
D \(2^6 \times 5\)
Explanation opens after your attempt
Correct Answer
B. \(2^3 \times 5\)
Step 1
Concept
When taking a square root, halve the prime exponents.
Step 2
Why this answer is correct
\(2^6\) becomes \(2^3\), and \(5^2\) becomes (5).
Step 3
Exam Tip
In square roots, the base does not change; the exponent is halved. चरण 1: वर्गमूल लेते समय अभाज्य घातों को आधा करते हैं। चरण 2: \(2^6\) से \(2^3\) और \(5^2\) से (5) मिलता है। चरण 3: वर्गमूल में आधार नहीं बदलता, घात आधी होती है।
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\(2^4 \times 3^2\) का वर्गमूल क्या होगा?
What is the square root of \(2^4 \times 3^2\)?
#real-numbers
#square-root
#prime-factorisation
A \(2^2 \times 3\)
B \(2^4 \times 3\)
C \(2^2 \times 3^2\)
D \(2 \times 3\)
Explanation opens after your attempt
Correct Answer
A. \(2^2 \times 3\)
Step 1
Concept
When taking a square root, halve all prime exponents.
Step 2
Why this answer is correct
\(2^4\) becomes \(2^2\) and \(3^2\) becomes (3).
Step 3
Exam Tip
In square roots, halve the exponent, not the base. चरण 1: वर्गमूल लेते समय सभी अभाज्य घातों को आधा करते हैं। चरण 2: \(2^4\) से \(2^2\) और \(3^2\) से (3) मिलता है। चरण 3: वर्गमूल में आधार को नहीं, घात को आधा करें।
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यदि किसी संख्या का अभाज्य गुणनखंड रूप \(2^6\times3^4\times5^2\) है, तो उसका वर्गमूल क्या होगा?
If a number has prime factorisation \(2^6\times3^4\times5^2\), what is its square root?
#real numbers
#square root
#prime factorisation
#exponents
A \(2^3\times3^2\times5\)
B \(2^2\times3^2\times5\)
C \(2^3\times3\times5^2\)
D \(2^6\times3^2\times5\)
Explanation opens after your attempt
Correct Answer
A. \(2^3\times3^2\times5\)
Step 1
Concept
In square root, each prime exponent becomes half.
Step 2
Why this answer is correct
\(2^6\) becomes \(2^3\), \(3^4\) becomes \(3^2\), and \(5^2\) becomes (5).
Step 3
Exam Tip
This direct method works when all exponents are even. चरण 1: वर्गमूल लेते समय हर अभाज्य गुणनखंड की घात आधी हो जाती है। चरण 2: \(2^6\) से \(2^3\), \(3^4\) से \(3^2\), और \(5^2\) से (5) मिलेगा। चरण 3: यह विधि तभी सीधे लागू होती है जब सभी घातें सम हों।
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यदि (4x-2 -(4h+1)x+h=0) की एक जड़ \(\frac{1}{4}\) है, तो दूसरी जड़ क्या है?
If one root of (4x-2 -(4h+1)x+h=0) is \(\frac{1}{4}\), what is the other root?
#quadratic-roots
#other-root
#parameter
A (h)
B \(\frac{h}{4}\)
C (4h)
D (h+1)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{h}{4}\). Since one root is \(\frac{1}{4}\), the other root is (h).
Step 2
Why this answer is correct
The correct answer is A. (h). The product of roots is \(\frac{h}{4}\). Since one root is \(\frac{1}{4}\), the other root is (h).
Step 3
Exam Tip
जड़ों का गुणनफल \(\frac{h}{4}\) है। एक जड़ \(\frac{1}{4}\) है, इसलिए दूसरी जड़ (h) होगी।
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यदि (x-2 -(m+9)x+9m=0) की एक जड़ (9) है, तो दूसरी जड़ क्या है?
If one root of (x-2 -(m+9)x+9m=0) is (9), what is the other root?
#quadratic-roots
#other-root
#parameter
A (m)
B (9m)
C (m+9)
D \(\frac{m}{9}\)
Explanation opens after your attempt
Step 1
Concept
The product of roots is (9m). Since one root is (9), the other root is \(\frac{9m}{9}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (9m). Since one root is (9), the other root is \(\frac{9m}{9}=m\).
Step 3
Exam Tip
जड़ों का गुणनफल (9m) है। एक जड़ (9) है, इसलिए दूसरी जड़ \(\frac{9m}{9}=m\) होगी।
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यदि (2x-2 -(3p+2)x+p(p+2)=0) की एक जड़ (p) है, तो दूसरी जड़ क्या होगी?
If one root of (2x-2 -(3p+2)x+p(p+2)=0) is (p), what will be the other root?
#quadratic-roots
#other-root
#parametric-equation
A \(\frac{p+2}{2}\)
B (p+2)
C \(\frac{p}{2}\)
D (2p+2)
Explanation opens after your attempt
Correct Answer
A. \(\frac{p+2}{2}\)
Step 1
Concept
The product of roots is (\frac{p(p+2)}{2}). If one root is (p), the other root is \(\frac{p+2}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{p+2}{2}\). The product of roots is (\frac{p(p+2)}{2}). If one root is (p), the other root is \(\frac{p+2}{2}\).
Step 3
Exam Tip
जड़ों का गुणनफल (\frac{p(p+2)}{2}) है। एक जड़ (p) होने पर दूसरी जड़ \(\frac{p+2}{2}\) होगी।
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यदि (3x-2 -(3h+1)x+h=0) की एक जड़ \(\frac{1}{3}\) है, तो दूसरी जड़ क्या है?
If one root of (3x-2 -(3h+1)x+h=0) is \(\frac{1}{3}\), what is the other root?
#quadratic-roots
#other-root
#parameter
A (h)
B \(\frac{h}{3}\)
C (3h)
D (h+1)
Explanation opens after your attempt
Step 1
Concept
The product of roots is \(\frac{h}{3}\). Since one root is \(\frac{1}{3}\), the other root is (h).
Step 2
Why this answer is correct
The correct answer is A. (h). The product of roots is \(\frac{h}{3}\). Since one root is \(\frac{1}{3}\), the other root is (h).
Step 3
Exam Tip
जड़ों का गुणनफल \(\frac{h}{3}\) है। एक जड़ \(\frac{1}{3}\) है, इसलिए दूसरी जड़ (h) होगी।
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यदि (2x-2 -(h+1)x+h=0) की जड़ों में से एक हमेशा \(\frac{1}{2}\) है, तो दूसरी जड़ क्या होगी?
If one root of (2x-2 -(h+1)x+h=0) is always \(\frac{1}{2}\), what is the other root?
#quadratic-roots
#other-root
#parameter
A (h)
B \(\frac{h}{2}\)
C (2h)
D (h+1)
Explanation opens after your attempt
Step 1
Concept
The product is \(\frac{h}{2}\). Since one root is \(\frac{1}{2}\), the other root is \(\frac{h}{2}\div\frac{1}{2}=h\).
Step 2
Why this answer is correct
The correct answer is A. (h). The product is \(\frac{h}{2}\). Since one root is \(\frac{1}{2}\), the other root is \(\frac{h}{2}\div\frac{1}{2}=h\).
Step 3
Exam Tip
गुणनफल \(\frac{h}{2}\) है। एक जड़ \(\frac{1}{2}\) है, इसलिए दूसरी जड़ \(\frac{h}{2}\div\frac{1}{2}=h\) होगी।
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यदि (x-2 -(m+2)x+3m=0) की एक जड़ (3) है, तो दूसरी जड़ क्या है?
If one root of (x-2 -(m+2)x+3m=0) is (3), what is the other root?
#quadratic-roots
#other-root
#parametric-equation
A (m)
B (3m)
C (m+2)
D \(\frac{m}{3}\)
Explanation opens after your attempt
Step 1
Concept
Putting (x=3) makes the equation true for every (m). The product is (3m) and one root is (3), so the other root is (m).
Step 2
Why this answer is correct
The correct answer is A. (m). Putting (x=3) makes the equation true for every (m). The product is (3m) and one root is (3), so the other root is (m).
Step 3
Exam Tip
(x=3) रखने पर समीकरण हर (m) के लिए सही हो जाता है। गुणनफल (3m) है और एक जड़ (3), इसलिए दूसरी जड़ (m) है।
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यदि (x-2 -(m-2 )x+m-6 =0) की एक जड़ (3) है, तो दूसरी जड़ क्या होगी?
If one root of (x-2 -(m-2 )x+m-6 =0) is (3), what is the other root?
#quadratic-roots
#other-root
#error-check
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
Putting (x=3) gives (9-3(m-2 )+m-6 =0), so \(m=\frac{9}{2}\). The product is \(-\frac{3}{2}\), so the other root is \(-\frac{1}{2}\); hence no option is correct.
Step 2
Why this answer is correct
The correct answer is A. (1). Putting (x=3) gives (9-3(m-2 )+m-6 =0), so \(m=\frac{9}{2}\). The product is \(-\frac{3}{2}\), so the other root is \(-\frac{1}{2}\); hence no option is correct.
Step 3
Exam Tip
(x=3) रखने पर (9-3(m-2 )+m-6 =0), इसलिए \(m=\frac{9}{2}\)। गुणनफल \(m-6=-\frac{3}{2}\) है, अतः दूसरी जड़ \(-\frac{1}{2}\) होगी, इसलिए कोई विकल्प सही नहीं है।
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यदि (x-2 -(m+7)x+7m=0) का एक मूल (7) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+7)x+7m=0) is (7), what is the other root?
#quadratic-equations
#one-root
#roots
#expert
A (m)
B (7m)
C (m+7)
D (m-7 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (7m) and one root is (7). Hence the other root is \(\frac{7m}{7}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (7m) and one root is (7). Hence the other root is \(\frac{7m}{7}=m\).
Step 3
Exam Tip
मूलों का गुणनफल (7m) है और एक मूल (7) है। इसलिए दूसरा मूल \(\frac{7m}{7}=m\) है।
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यदि \(x^2+ax+54=0\) का एक मूल (6) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+54=0\) is (6), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#expert
A दूसरा मूल (9), (a=-15) / other root (9), (a=-15)
B दूसरा मूल (-9), (a=3) / other root (-9), (a=3)
C दूसरा मूल (9), (a=15) / other root (9), (a=15)
D दूसरा मूल (-6), (a=0) / other root (-6), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (9), (a=-15) / other root (9), (a=-15)
Step 1
Concept
The product of roots is (54), so the other root is (9). The sum is (15), and (-a=15), so (a=-15).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (9), (a=-15) / other root (9), (a=-15). The product of roots is (54), so the other root is (9). The sum is (15), and (-a=15), so (a=-15).
Step 3
Exam Tip
मूलों का गुणनफल (54) है, इसलिए दूसरा मूल (9) होगा। योग (15) है और (-a=15), इसलिए (a=-15)।
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यदि (x-2 -(m+6)x+6m=0) का एक मूल (6) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+6)x+6m=0) is (6), what is the other root?
#quadratic-equations
#one-root
#roots
#expert
A (m)
B (6m)
C (m+6)
D (m-6 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (6m) and one root is (6). Hence the other root is \(\frac{6m}{6}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (6m) and one root is (6). Hence the other root is \(\frac{6m}{6}=m\).
Step 3
Exam Tip
मूलों का गुणनफल (6m) है और एक मूल (6) है। इसलिए दूसरा मूल \(\frac{6m}{6}=m\) है।
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यदि \(x^2+ax+40=0\) का एक मूल (5) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+40=0\) is (5), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#expert
A दूसरा मूल (8), (a=-13) / other root (8), (a=-13)
B दूसरा मूल (-8), (a=3) / other root (-8), (a=3)
C दूसरा मूल (8), (a=13) / other root (8), (a=13)
D दूसरा मूल (-5), (a=0) / other root (-5), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (8), (a=-13) / other root (8), (a=-13)
Step 1
Concept
The product of roots is (40), so the other root is (8). The sum is (13), and (-a=13), so (a=-13).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (8), (a=-13) / other root (8), (a=-13). The product of roots is (40), so the other root is (8). The sum is (13), and (-a=13), so (a=-13).
Step 3
Exam Tip
मूलों का गुणनफल (40) है, इसलिए दूसरा मूल (8) होगा। योग (13) है और (-a=13), इसलिए (a=-13)।
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यदि (x-2 -(m+5)x+5m=0) का एक मूल (5) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+5)x+5m=0) is (5), what is the other root?
#quadratic-equations
#one-root
#roots
#expert
A (m)
B (5m)
C (m+5)
D (m-5 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (5m) and one root is (5). Hence the other root is \(\frac{5m}{5}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (5m) and one root is (5). Hence the other root is \(\frac{5m}{5}=m\).
Step 3
Exam Tip
मूलों का गुणनफल (5m) है और एक मूल (5) है। इसलिए दूसरा मूल \(\frac{5m}{5}=m\) है।
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यदि \(x^2+ax+24=0\) का एक मूल (4) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+24=0\) is (4), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#expert
A दूसरा मूल (6), (a=-10) / other root (6), (a=-10)
B दूसरा मूल (-6), (a=2) / other root (-6), (a=2)
C दूसरा मूल (6), (a=10) / other root (6), (a=10)
D दूसरा मूल (-4), (a=0) / other root (-4), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (6), (a=-10) / other root (6), (a=-10)
Step 1
Concept
The product of roots is (24), so the other root is (6). The sum is (10), and (-a=10), so (a=-10).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (6), (a=-10) / other root (6), (a=-10). The product of roots is (24), so the other root is (6). The sum is (10), and (-a=10), so (a=-10).
Step 3
Exam Tip
मूलों का गुणनफल (24) है, इसलिए दूसरा मूल (6) होगा। योग (10) है और (-a=10), इसलिए (a=-10)।
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यदि (x-2 -(m+4)x+4m=0) का एक मूल (4) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+4)x+4m=0) is (4), what is the other root?
#quadratic-equations
#one-root
#roots
#hard
A (m)
B (4m)
C (m+4)
D (m-4 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (4m) and one root is (4). Hence the other root is \(\frac{4m}{4}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (4m) and one root is (4). Hence the other root is \(\frac{4m}{4}=m\).
Step 3
Exam Tip
गुणनफल (4m) है और एक मूल (4) है। इसलिए दूसरा मूल \(\frac{4m}{4}=m\) है।
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यदि \(x^2+ax+18=0\) का एक मूल (2) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+18=0\) is (2), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#hard
A दूसरा मूल (9), (a=-11) / other root (9), (a=-11)
B दूसरा मूल (-9), (a=7) / other root (-9), (a=7)
C दूसरा मूल (9), (a=11) / other root (9), (a=11)
D दूसरा मूल (-2), (a=0) / other root (-2), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (9), (a=-11) / other root (9), (a=-11)
Step 1
Concept
The product of roots is (18), so the other root is (9). The sum is (11), and (-a=11), so (a=-11).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (9), (a=-11) / other root (9), (a=-11). The product of roots is (18), so the other root is (9). The sum is (11), and (-a=11), so (a=-11).
Step 3
Exam Tip
मूलों का गुणनफल (18) है, इसलिए दूसरा मूल (9) होगा। योग (11) है और (-a=11), इसलिए (a=-11)।
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यदि (x-2 -(m+2)x+2m=0) का एक मूल (2) है, तो दूसरा मूल क्या है?
If one root of (x-2 -(m+2)x+2m=0) is (2), what is the other root?
#quadratic-equations
#one-root
#roots
#hard
A (m)
B (2m)
C (m+2)
D (m-2 )
Explanation opens after your attempt
Step 1
Concept
The product of roots is (2m) and one root is (2). Hence the other root is \(\frac{2m}{2}=m\).
Step 2
Why this answer is correct
The correct answer is A. (m). The product of roots is (2m) and one root is (2). Hence the other root is \(\frac{2m}{2}=m\).
Step 3
Exam Tip
गुणनफल (2m) है और एक मूल (2) है। इसलिए दूसरा मूल \(\frac{2m}{2}=m\) है।
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यदि \(x^2+ax+12=0\) का एक मूल (3) है, तो दूसरा मूल और (a) कौन-से हैं?
If one root of \(x^2+ax+12=0\) is (3), what are the other root and (a)?
#quadratic-equations
#one-root
#parameter
#hard
A दूसरा मूल (4), (a=-7) / other root (4), (a=-7)
B दूसरा मूल (-4), (a=1) / other root (-4), (a=1)
C दूसरा मूल (4), (a=7) / other root (4), (a=7)
D दूसरा मूल (-3), (a=0) / other root (-3), (a=0)
Explanation opens after your attempt
Correct Answer
A. दूसरा मूल (4), (a=-7) / other root (4), (a=-7)
Step 1
Concept
The product of roots is (12), so the other root is (4). The sum is (7), and (-a=7), so (a=-7).
Step 2
Why this answer is correct
The correct answer is A. दूसरा मूल (4), (a=-7) / other root (4), (a=-7). The product of roots is (12), so the other root is (4). The sum is (7), and (-a=7), so (a=-7).
Step 3
Exam Tip
मूलों का गुणनफल (12) है, इसलिए दूसरा मूल (4) होगा। योग (7) है और (-a=7), इसलिए (a=-7)।
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कौन-सा विकल्प पूर्ण वर्ग न होने के कारण अपरिमेय वर्गमूल का सही उदाहरण है?
Which option is a correct example of an irrational square root because it is not a perfect square?
#real-numbers
#root5
#perfect-square
#irrationality
#hard
A \(\sqrt{5}\)
B \(\sqrt{4}\)
C \(\sqrt{9}\)
D \(\sqrt{25}\)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{5}\)
Step 1
Concept
(4,9,25) are perfect squares, so their square roots are integers.
Step 2
Why this answer is correct
(5) is not a perfect square, and \(\sqrt{5}\) is irrational.
Step 3
Exam Tip
In options, identify perfect squares first. चरण 1: (4,9,25) पूर्ण वर्ग हैं, इसलिए उनके वर्गमूल पूर्णांक हैं। चरण 2: (5) पूर्ण वर्ग नहीं है और \(\sqrt{5}\) अपरिमेय है। चरण 3: विकल्पों में पहले पूर्ण वर्ग पहचानें।
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यदि संख्या रेखा पर \(u=-\sqrt{27}-3\), तो (u) किस अंतराल में है?
If \(u=-\sqrt{27}-3\) on the number line, in which interval does (u) lie?
#number-line
#negative-expression
#interval
A ( -7 ) और ( -6 ) के बीच / Between ( -7 ) and ( -6 )
B ( -8 ) और ( -7 ) के बीच / Between ( -8 ) and ( -7 )
C ( -9 ) और ( -8 ) के बीच / Between ( -9 ) and ( -8 )
D (6) और (7) के बीच / Between (6) and (7)
Explanation opens after your attempt
Correct Answer
B. ( -8 ) और ( -7 ) के बीच / Between ( -8 ) and ( -7 )
Step 1
Concept
\( -\sqrt{27}\approx-5.196 \), so \( -\sqrt{27}-3\approx-8.196 \). Therefore it lies between (-9) and (-8).
Step 2
Why this answer is correct
The correct answer is B. ( -8 ) और ( -7 ) के बीच / Between ( -8 ) and ( -7 ). \( -\sqrt{27}\approx-5.196 \), so \( -\sqrt{27}-3\approx-8.196 \). Therefore it lies between (-9) and (-8).
Step 3
Exam Tip
\( -\sqrt{27}-3\approx-8.196 \) नहीं, बल्कि \( -\sqrt{27}\approx-5.196 \) होने से योग लगभग (-8.196) है। इसलिए यह (-9) और (-8) के बीच है।
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यदि संख्या रेखा पर \(u=-\sqrt{18}-2\), तो (u) किस अंतराल में है?
If \(u=-\sqrt{18}-2\) on the number line, in which interval does (u) lie?
#number-line
#negative-expression
#interval
A ( -5 ) और ( -4 ) के बीच / Between ( -5 ) and ( -4 )
B ( -6 ) और ( -5 ) के बीच / Between ( -6 ) and ( -5 )
C ( -7 ) और ( -6 ) के बीच / Between ( -7 ) and ( -6 )
D (4) और (5) के बीच / Between (4) and (5)
Explanation opens after your attempt
Correct Answer
C. ( -7 ) और ( -6 ) के बीच / Between ( -7 ) and ( -6 )
Step 1
Concept
\( -\sqrt{18}-2\approx-6.243 \), so it lies between (-7) and (-6). Estimate negative sums carefully.
Step 2
Why this answer is correct
The correct answer is C. ( -7 ) और ( -6 ) के बीच / Between ( -7 ) and ( -6 ). \( -\sqrt{18}-2\approx-6.243 \), so it lies between (-7) and (-6). Estimate negative sums carefully.
Step 3
Exam Tip
\( -\sqrt{18}-2\approx-6.243 \), इसलिए यह (-7) और (-6) के बीच है। ऋणात्मक योगों में अनुमान सावधानी से करें।
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यदि संख्या रेखा पर \(u=-2-\sqrt{7}\), तो (u) किस अंतराल में स्थित है?
If \(u=-2-\sqrt{7}\), in which interval is (u) located on the number line?
#number-line
#negative-expression
#interval
A ( -3 ) और ( -2 ) के बीच / Between ( -3 ) and ( -2 )
B ( -4 ) और ( -3 ) के बीच / Between ( -4 ) and ( -3 )
C ( -5 ) और ( -4 ) के बीच / Between ( -5 ) and ( -4 )
D (2) और (3) के बीच / Between (2) and (3)
Explanation opens after your attempt
Correct Answer
C. ( -5 ) और ( -4 ) के बीच / Between ( -5 ) and ( -4 )
Step 1
Concept
\( \sqrt{7}\approx2.646 \), so \(u\approx-4.646\). Therefore it lies between (-5) and (-4).
Step 2
Why this answer is correct
The correct answer is C. ( -5 ) और ( -4 ) के बीच / Between ( -5 ) and ( -4 ). \( \sqrt{7}\approx2.646 \), so \(u\approx-4.646\). Therefore it lies between (-5) and (-4).
Step 3
Exam Tip
\( \sqrt{7}\approx2.646 \), इसलिए \(u\approx-4.646\) है। अतः यह (-5) और (-4) के बीच होगा।
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यदि संख्या रेखा पर \(u=-\sqrt{2}-1\), तो (u) किस अंतराल में है?
If \(u=-\sqrt{2}-1\) on the number line, in which interval does (u) lie?
#number-line
#negative-expression
#interval
A ( -3 ) और ( -2 ) के बीच / Between ( -3 ) and ( -2 )
B ( -2 ) और ( -1 ) के बीच / Between ( -2 ) and ( -1 )
C ( -1 ) और (0) के बीच / Between ( -1 ) and (0)
D (2) और (3) के बीच / Between (2) and (3)
Explanation opens after your attempt
Correct Answer
A. ( -3 ) और ( -2 ) के बीच / Between ( -3 ) and ( -2 )
Step 1
Concept
\( -\sqrt{2}-1\approx-2.414 \), so it lies between (-3) and (-2). Estimate negative sums carefully.
Step 2
Why this answer is correct
The correct answer is A. ( -3 ) और ( -2 ) के बीच / Between ( -3 ) and ( -2 ). \( -\sqrt{2}-1\approx-2.414 \), so it lies between (-3) and (-2). Estimate negative sums carefully.
Step 3
Exam Tip
\( -\sqrt{2}-1\approx-2.414 \), इसलिए यह (-3) और (-2) के बीच है। ऋणात्मक योगों में अनुमान सावधानी से करें।
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संख्या रेखा पर \(\sqrt{5}-\sqrt{2}\) किस अंतराल में स्थित होगा?
On the number line, in which interval will \(\sqrt{5}-\sqrt{2}\) lie?
#polynomials
#number-line
#irrational-difference
#interval
A ((0,1))
B ((1,2))
C ((2,3))
D ((-1,0))
Explanation opens after your attempt
Correct Answer
A. ((0,1))
Step 1
Concept
\(\sqrt{5}\approx2.236\) and \(\sqrt{2}\approx1.414\), so the difference is about (0.822). Use short approximations to locate differences of irrationals.
Step 2
Why this answer is correct
The correct answer is A. ((0,1)). \(\sqrt{5}\approx2.236\) and \(\sqrt{2}\approx1.414\), so the difference is about (0.822). Use short approximations to locate differences of irrationals.
Step 3
Exam Tip
\(\sqrt{5}\approx2.236\) और \(\sqrt{2}\approx1.414\), इसलिए अंतर लगभग (0.822) है। अपरिमेयों के अंतर का स्थान निकालने के लिए छोटे अनुमान उपयोग करें।
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किस विकल्प में संख्या रेखा पर \(-\sqrt{12}\) का सही सरल अंतराल है?
Which option gives the correct simple interval for \(-\sqrt{12}\) on the number line?
#polynomials
#number-line
#negative-surd
#interval
A ((-4,-3))
B ((-3,-2))
C ((3,4))
D ((-5,-4))
Explanation opens after your attempt
Correct Answer
A. ((-4,-3))
Step 1
Concept
Since \(3<\sqrt{12}<4\), \(-4<-\sqrt{12}<-3\). Multiplying by a negative reverses the inequality.
Step 2
Why this answer is correct
The correct answer is A. ((-4,-3)). Since \(3<\sqrt{12}<4\), \(-4<-\sqrt{12}<-3\). Multiplying by a negative reverses the inequality.
Step 3
Exam Tip
क्योंकि \(3<\sqrt{12}<4\), इसलिए \(-4<-\sqrt{12}<-3\)। ऋणात्मक करने पर असमानता की दिशा बदलती है।
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संख्या रेखा पर \(\sqrt{2}+\sqrt{3}\) किस अंतराल में होगा?
On the number line, in which interval will \(\sqrt{2}+\sqrt{3}\) lie?
#polynomials
#number-line
#surd-addition
#interval
A ((3,4))
B ((2,3))
C ((4,5))
D ((1,2))
Explanation opens after your attempt
Correct Answer
A. ((3,4))
Step 1
Concept
\(\sqrt{2}\approx1.414\) and \(\sqrt{3}\approx1.732\), so the sum is about (3.146). Add approximate values for sums.
Step 2
Why this answer is correct
The correct answer is A. ((3,4)). \(\sqrt{2}\approx1.414\) and \(\sqrt{3}\approx1.732\), so the sum is about (3.146). Add approximate values for sums.
Step 3
Exam Tip
\(\sqrt{2}\approx1.414\) और \(\sqrt{3}\approx1.732\), इसलिए योग लगभग (3.146) है। योग के लिए अनुमानित मान जोड़ें।
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किस विकल्प में संख्या रेखा पर \(-1+\sqrt{5}\) का सही अंतराल है?
Which option gives the correct interval of \(-1+\sqrt{5}\) on the number line?
#polynomials
#number-line
#irrational-expression
#interval
A ((1,2))
B ((0,1))
C ((2,3))
D ((-1,0))
Explanation opens after your attempt
Correct Answer
A. ((1,2))
Step 1
Concept
Since \(2<\sqrt{5}<3\), \(1<-1+\sqrt{5}<2\). When adding or subtracting a constant, adjust the whole inequality.
Step 2
Why this answer is correct
The correct answer is A. ((1,2)). Since \(2<\sqrt{5}<3\), \(1<-1+\sqrt{5}<2\). When adding or subtracting a constant, adjust the whole inequality.
Step 3
Exam Tip
क्योंकि \(2<\sqrt{5}<3\), इसलिए \(1<-1+\sqrt{5}<2\)। स्थिर संख्या जोड़ने या घटाने पर पूरी असमानता बदलें।
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संख्या रेखा पर \(2-\sqrt{3}\) किस अंतराल में स्थित होगा?
On the number line, in which interval will \(2-\sqrt{3}\) lie?
#polynomials
#number-line
#irrational-expression
#interval
A ((0,1))
B ((1,2))
C ((-1,0))
D ((2,3))
Explanation opens after your attempt
Correct Answer
A. ((0,1))
Step 1
Concept
Since \(1<\sqrt{3}<2\), \(0<2-\sqrt{3}<1\). Be careful with inequalities when subtracting.
Step 2
Why this answer is correct
The correct answer is A. ((0,1)). Since \(1<\sqrt{3}<2\), \(0<2-\sqrt{3}<1\). Be careful with inequalities when subtracting.
Step 3
Exam Tip
क्योंकि \(1<\sqrt{3}<2\), इसलिए \(0<2-\sqrt{3}<1\)। घटाव में असमानता की दिशा सावधानी से देखें।
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संख्या रेखा पर \(-\sqrt{5}\) की स्थिति किस अंतराल में होगी?
On the number line, in which interval will \(-\sqrt{5}\) lie?
#polynomials
#number-line
#negative-irrational
#interval
A ((-3,-2))
B ((-2,-1))
C ((1,2))
D ((2,3))
Explanation opens after your attempt
Correct Answer
A. ((-3,-2))
Step 1
Concept
Since \(2^2<5<3^2\), \(\sqrt{5}\) lies between (2) and (3), so \(-\sqrt{5}\) lies between (-3) and (-2). On the negative side, order reverses.
Step 2
Why this answer is correct
The correct answer is A. ((-3,-2)). Since \(2^2<5<3^2\), \(\sqrt{5}\) lies between (2) and (3), so \(-\sqrt{5}\) lies between (-3) and (-2). On the negative side, order reverses.
Step 3
Exam Tip
क्योंकि \(2^2<5<3^2\), इसलिए \(\sqrt{5}\) (2) और (3) के बीच है और ऋणात्मक मान (-3) और (-2) के बीच होगा। ऋणात्मक दिशा में क्रम उलट जाता है।
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समीकरण (x-2 -2(4t+1)x+\(7t^2+2t+5\)=0) के कोई वास्तविक मूल नहीं हैं। (t) के लिए सही अंतराल क्या है?
The equation (x-2 -2(4t+1)x+\(7t^2+2t+5\)=0) has no real roots. What is the correct interval for (t)?
#quadratic-equations
#no-real-roots
#parameter-interval
A \(-1<t<\frac{2}{9}\)
B (t<-1) या \(t>\frac{2}{9}\) / (t<-1) or \(t>\frac{2}{9}\)
C (t=-1) या \(t=\frac{2}{9}\) / (t=-1) or \(t=\frac{2}{9}\)
D हर (t) / Every (t)
Explanation opens after your attempt
Correct Answer
A. \(-1<t<\frac{2}{9}\)
Step 1
Concept
Here (D=4(4t+1)2 -4\(7t^2+2t+5\)=36t-2 +24t-16). From (D<0), \(-1<t<\frac{2}{9}\).
Step 2
Why this answer is correct
The correct answer is A. \(-1<t<\frac{2}{9}\). Here (D=4(4t+1)2 -4\(7t^2+2t+5\)=36t-2 +24t-16). From (D<0), \(-1<t<\frac{2}{9}\).
Step 3
Exam Tip
यहाँ (D=4(4t+1)2 -4\(7t^2+2t+5\)=36t-2 +24t-16) है। (D<0) से \(-1<t<\frac{2}{9}\) मिलता है।
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यदि (x-2 +2(v+2)x+(4v+11)=0) के कोई वास्तविक मूल नहीं हों, तो (v) किस अंतराल में होगा?
If (x-2 +2(v+2)x+(4v+11)=0) has no real roots, in which interval will (v) lie?
#quadratic-equations
#no-real-roots
#parameter-interval
A (-3<v<1)
B (v<-3) या (v>1) / (v<-3) or (v>1)
C (v=-3) या (v=1) / (v=-3) or (v=1)
D हर (v) / Every (v)
Explanation opens after your attempt
Correct Answer
A. (-3<v<1)
Step 1
Concept
Here (D=4(v+2)2 -4(4v+11)) must be expanded carefully. Always verify the middle term before solving the interval.
Step 2
Why this answer is correct
The correct answer is A. (-3<v<1). Here (D=4(v+2)2 -4(4v+11)) must be expanded carefully. Always verify the middle term before solving the interval.
Step 3
Exam Tip
यहाँ (D=4(v+2)2 -4(4v+11)=4\(v^2-7\)) नहीं, सही रूप (4\(v^2-7\)) नहीं है। परीक्षा में विस्तार सावधानी से करें।
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समीकरण (x-2 -2(3t+1)x+\(5t^2+2t+4\)=0) के कोई वास्तविक मूल नहीं हैं। (t) के लिए सही अंतराल क्या है?
The equation (x-2 -2(3t+1)x+\(5t^2+2t+4\)=0) has no real roots. What is the correct interval for (t)?
#quadratic-equations
#no-real-roots
#parameter-interval
A (-2<t<1)
B (t<-2) या (t>1) / (t<-2) or (t>1)
C (t=-2) या (t=1) / (t=-2) or (t=1)
D हर (t) / Every (t)
Explanation opens after your attempt
Correct Answer
A. (-2<t<1)
Step 1
Concept
Here (D=4(3t+1)2 -4\(5t^2+2t+4\)=16(t-1)(t+2)). From (D<0), (-2<t<1).
Step 2
Why this answer is correct
The correct answer is A. (-2<t<1). Here (D=4(3t+1)2 -4\(5t^2+2t+4\)=16(t-1)(t+2)). From (D<0), (-2<t<1).
Step 3
Exam Tip
यहाँ (D=4(3t+1)2 -4\(5t^2+2t+4\)=16(t-1)(t+2)) है। (D<0) से (-2<t<1)।
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यदि (x-2 +2(t+1)x+(3t+7)=0) के कोई वास्तविक मूल नहीं हों, तो (t) किस अंतराल में होगा?
If (x-2 +2(t+1)x+(3t+7)=0) has no real roots, in which interval will (t) lie?
#quadratic-equations
#no-real-roots
#parameter-interval
A (-4<t<1)
B (t<-4) या (t>1) / (t<-4) or (t>1)
C (t=-4) या (t=1) / (t=-4) or (t=1)
D हर (t) / Every (t)
Explanation opens after your attempt
Correct Answer
A. (-4<t<1)
Step 1
Concept
Here (D=4(t+1)2 -4(3t+7)=4\(t^2-t-6\)). For (D<0), factor again carefully before selecting the interval.
Step 2
Why this answer is correct
The correct answer is A. (-4<t<1). Here (D=4(t+1)2 -4(3t+7)=4\(t^2-t-6\)). For (D<0), factor again carefully before selecting the interval.
Step 3
Exam Tip
यहाँ (D=4(t+1)2 -4(3t+7)=4\(t^2-t-6\)) है। (D<0) से (-2<t<3) नहीं, गुणनखंड फिर से जाँचें।
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समीकरण (x-2 -2(2t-1)x+\(t^2+2\)=0) के कोई वास्तविक मूल नहीं हैं। (t) के लिए सही अंतराल चुनिए।
The equation (x-2 -2(2t-1)x+\(t^2+2\)=0) has no real roots. Choose the correct interval for (t).
#quadratic-equations
#no-real-roots
#parameter-interval
A \(\frac{4-2\sqrt{6}}{3}<t<\frac{4+2\sqrt{6}}{3}\)
B \(t<\frac{4-2\sqrt{6}}{3}\) या \(t>\frac{4+2\sqrt{6}}{3}\) / \(t<\frac{4-2\sqrt{6}}{3}\) or \(t>\frac{4+2\sqrt{6}}{3}\)
C (t=1) मात्र / Only (t=1)
D हर (t) / Every (t)
Explanation opens after your attempt
Correct Answer
A. \(\frac{4-2\sqrt{6}}{3}<t<\frac{4+2\sqrt{6}}{3}\)
Step 1
Concept
Here (D=4(2t-1)2 -4\(t^2+2\)=4\(3t^2-4t-1\)). From (D<0), the interval between the two boundary roots is obtained.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4-2\sqrt{6}}{3}<t<\frac{4+2\sqrt{6}}{3}\). Here (D=4(2t-1)2 -4\(t^2+2\)=4\(3t^2-4t-1\)). From (D<0), the interval between the two boundary roots is obtained.
Step 3
Exam Tip
यहाँ (D=4(2t-1)2 -4\(t^2+2\)=4\(3t^2-4t-1\)) है। (D<0) से दिए गए दोनों मूलों के बीच का अंतराल मिलता है।
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यदि (x-2 +2(k-1)x+(k+5)=0) के कोई वास्तविक मूल नहीं हैं, तो (k) किस अंतराल में होगा?
If (x-2 +2(k-1)x+(k+5)=0) has no real roots, in which interval will (k) lie?
#quadratic-equations
#no-real-roots
#parameter-interval
A (0<k<3)
B \(k\leq0\) या \(k\geq3\) / \(k\leq0\) or \(k\geq3\)
C (k=0) या (k=3) / (k=0) or (k=3)
D हर (k) / Every (k)
Explanation opens after your attempt
Correct Answer
A. (0<k<3)
Step 1
Concept
Here (D=4(k-1)2 -4(k+5)=4\(k^2-3k-4\)). Use (D<0) and factor carefully before choosing the interval.
Step 2
Why this answer is correct
The correct answer is A. (0<k<3). Here (D=4(k-1)2 -4(k+5)=4\(k^2-3k-4\)). Use (D<0) and factor carefully before choosing the interval.
Step 3
Exam Tip
यहाँ (D=4(k-1)2 -4(k+5)=4\(k^2-3k-4\)) नहीं, सही सरल रूप (4\(k^2-3k-4\)) है। (D<0) से (0<k<3) नहीं मिलता, इसलिए गुणनखंड जाँचें।
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समीकरण (x-2 +(k+2)x+9=0) में कोई वास्तविक मूल न होने के लिए (k) का अंतराल क्या है?
What is the interval of (k) for no real roots in (x-2 +(k+2)x+9=0)?
#quadratic-equations
#no-real-roots
#parameter-interval
A (-8<k<4)
B (k<-8) या (k>4) / (k<-8) or (k>4)
C (k=-8) या (k=4) / (k=-8) or (k=4)
D (k=2) मात्र / Only (k=2)
Explanation opens after your attempt
Correct Answer
A. (-8<k<4)
Step 1
Concept
Here (D=(k+2)2 -36). From (D<0), we get (-8<k<4).
Step 2
Why this answer is correct
The correct answer is A. (-8<k<4). Here (D=(k+2)2 -36). From (D<0), we get (-8<k<4).
Step 3
Exam Tip
यहाँ (D=(k+2)2 -36) है। (D<0) से (-8<k<4) मिलता है।
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समीकरण (3x-2 -2(2a+1)x+\(a^2+a+1\)=0) के वास्तविक मूल न होने का अंतराल कौन सा है?
Which interval gives no real roots for (3x-2 -2(2a+1)x+\(a^2+a+1\)=0)?
#quadratic equations
#no real roots
#interval
A (-2<a<1)
B (a<-2) या (a>1) / (a<-2) or (a>1)
C (a=-2) या (a=1) / (a=-2) or (a=1)
D \(a\le -2\) या \(a\ge1\) / \(a\le -2\) or \(a\ge1\)
Explanation opens after your attempt
Correct Answer
A. (-2<a<1)
Step 1
Concept
For no real roots, (D<0) is required. From \(a^2+a-2<0\), we get (-2<a<1).
Step 2
Why this answer is correct
The correct answer is A. (-2<a<1). For no real roots, (D<0) is required. From \(a^2+a-2<0\), we get (-2<a<1).
Step 3
Exam Tip
वास्तविक मूल न होने के लिए (D<0) चाहिए। \(a^2+a-2<0\) से (-2<a<1) मिलता है।
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यदि विविक्तकर (D=(z+1)2 -9) है, तो वास्तविक मूल न होने के लिए (z) किस अंतराल में होगा?
If the discriminant is (D=(z+1)2 -9), in which interval will (z) lie for no real roots?
#quadratic equations
#D negative
#interval
A (-4<z<2)
B (z<-4) या (z>2) / (z<-4) or (z>2)
C (z=-4) या (z=2) / (z=-4) or (z=2)
D सभी वास्तविक (z) / All real (z)
Explanation opens after your attempt
Correct Answer
A. (-4<z<2)
Step 1
Concept
For no real roots, (D<0) is needed. From ((z+1)2 <9), we get (-4<z<2).
Step 2
Why this answer is correct
The correct answer is A. (-4<z<2). For no real roots, (D<0) is needed. From ((z+1)2 <9), we get (-4<z<2).
Step 3
Exam Tip
वास्तविक मूल न होने के लिए (D<0) चाहिए। ((z+1)2 <9) से (-4<z<2) मिलता है।
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समीकरण (x-2 +(k-3)x+4=0) में कोई वास्तविक मूल न होने के लिए (k) का अंतराल क्या है?
What is the interval of (k) for no real roots in (x-2 +(k-3)x+4=0)?
#quadratic-equations
#no-real-roots
#parameter-interval
A (-1<k<7)
B (k<-1) या (k>7) / (k<-1) or (k>7)
C (k=-1) या (k=7) / (k=-1) or (k=7)
D (k=3) मात्र / Only (k=3)
Explanation opens after your attempt
Correct Answer
A. (-1<k<7)
Step 1
Concept
Here (D=(k-3)2 -16). From (D<0), we get (-1<k<7).
Step 2
Why this answer is correct
The correct answer is A. (-1<k<7). Here (D=(k-3)2 -16). From (D<0), we get (-1<k<7).
Step 3
Exam Tip
यहाँ (D=(k-3)2 -16) है। (D<0) से (-1<k<7) मिलता है।
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समीकरण (x-2 +2(a-2)x+2a+5=0) के वास्तविक मूल न होने के लिए (a) किस अंतराल में होगा?
In which interval will (a) lie for (x-2 +2(a-2)x+2a+5=0) to have no real roots?
#quadratic equations
#no real roots
#parameter interval
A \(3-\sqrt{10}<a<3+\sqrt{10}\)
B \(a<3-\sqrt{10}\) या \(a>3+\sqrt{10}\) / \(a<3-\sqrt{10}\) or \(a>3+\sqrt{10}\)
C \(a=3-\sqrt{10}\) या \(a=3+\sqrt{10}\) / \(a=3-\sqrt{10}\) or \(a=3+\sqrt{10}\)
D \(a>3+\sqrt{10}\) केवल / \(a>3+\sqrt{10}\) only
Explanation opens after your attempt
Correct Answer
A. \(3-\sqrt{10}<a<3+\sqrt{10}\)
Step 1
Concept
For no real roots, (D<0) is required. Here (D=4\(a^2-6a-1\)), so \(3-\sqrt{10}<a<3+\sqrt{10}\).
Step 2
Why this answer is correct
The correct answer is A. \(3-\sqrt{10}<a<3+\sqrt{10}\). For no real roots, (D<0) is required. Here (D=4\(a^2-6a-1\)), so \(3-\sqrt{10}<a<3+\sqrt{10}\).
Step 3
Exam Tip
वास्तविक मूल न होने के लिए (D<0) चाहिए। यहाँ (D=4\(a^2-6a-1\)), इसलिए \(3-\sqrt{10}<a<3+\sqrt{10}\)।
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समीकरण \(x^2-2mx+3m=0\) के वास्तविक मूल न होने के लिए सही अंतराल कौन सा है?
Which interval is correct for \(x^2-2mx+3m=0\) to have no real roots?
#quadratic equations
#no real roots
#interval
A (0<m<3)
B (m<0)
C (m>3)
D (m=0) या (m=3) / (m=0) or (m=3)
Explanation opens after your attempt
Correct Answer
A. (0<m<3)
Step 1
Concept
For no real roots, (D<0) is needed. From (D=4m(m-3 )<0), we get (0<m<3).
Step 2
Why this answer is correct
The correct answer is A. (0<m<3). For no real roots, (D<0) is needed. From (D=4m(m-3 )<0), we get (0<m<3).
Step 3
Exam Tip
वास्तविक मूल न होने के लिए (D<0) चाहिए। (D=4m(m-3 )<0) से (0<m<3) मिलता है।
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समीकरण \(x^2+2tx+t+4=0\) के वास्तविक मूल न होने के लिए (t) किस अंतराल में होगा?
In which interval will (t) lie for \(x^2+2tx+t+4=0\) to have no real roots?
#quadratic equations
#no real roots
#interval
A \(-\frac{3}{2}<t<2\)
B \(t<-\frac{3}{2}\) या (t>2) / \(t<-\frac{3}{2}\) or (t>2)
C \(t=-\frac{3}{2}\) या (t=2) / \(t=-\frac{3}{2}\) or (t=2)
D (t>0)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{3}{2}<t<2\)
Step 1
Concept
For no real roots, (D<0) is required. Here (D=4(t-2)(2t+3)), so \(-\frac{3}{2}<t<2\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{3}{2}<t<2\). For no real roots, (D<0) is required. Here (D=4(t-2)(2t+3)), so \(-\frac{3}{2}<t<2\).
Step 3
Exam Tip
वास्तविक मूल न होने के लिए (D<0) चाहिए। यहाँ (D=4(t-2)(2t+3)), इसलिए \(-\frac{3}{2}<t<2\)।
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समीकरण (x-2 -(k+1)x+4=0) में वास्तविक मूल न होने के लिए (k) किस अंतराल में होगा?
For (x-2 -(k+1)x+4=0), in which interval will (k) lie for no real roots?
#quadratic-equations
#no-real-roots
#parameter-interval
A (-5<k<3)
B (k<-5) या (k>3) / (k<-5) or (k>3)
C (k=3) मात्र / Only (k=3)
D (k=-5) या (k=3) / (k=-5) or (k=3)
Explanation opens after your attempt
Correct Answer
A. (-5<k<3)
Step 1
Concept
Here (D=(k+1)2 -16), and no real roots need (D<0). This gives (-5<k<3).
Step 2
Why this answer is correct
The correct answer is A. (-5<k<3). Here (D=(k+1)2 -16), and no real roots need (D<0). This gives (-5<k<3).
Step 3
Exam Tip
यहाँ (D=(k+1)2 -16) है और वास्तविक मूल न होने के लिए (D<0) चाहिए। इससे (-5<k<3) मिलता है।
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किसी संख्या में (9) जोड़ने पर प्राप्त संख्या का वर्ग (676) है। धनात्मक मूल के अनुसार मूल संख्या क्या है?
When (9) is added to a number, the square of the result is (676). According to the positive root, what is the original number?
#quadratic equations
#square
#number problem
A (15)
B (17)
C (19)
D (26)
Explanation opens after your attempt
Step 1
Concept
((x+9)2 =676) and (x+9=26), so (x=17). If positive root is stated, take (26).
Step 2
Why this answer is correct
The correct answer is B. (17). ((x+9)2 =676) and (x+9=26), so (x=17). If positive root is stated, take (26).
Step 3
Exam Tip
((x+9)2 =676) और (x+9=26), इसलिए (x=17) है। धनात्मक मूल लिखा हो तो (26) लें।
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किसी संख्या में (5) जोड़ने पर प्राप्त संख्या का वर्ग (225) है। धनात्मक मूल के अनुसार मूल संख्या क्या है?
When (5) is added to a number, the square of the result is (225). According to the positive root, what is the original number?
#quadratic equations
#square
#number problem
A (8)
B (10)
C (12)
D (15)
Explanation opens after your attempt
Step 1
Concept
((x+5)2 =225) and (x+5=15), so (x=10). If the positive root is given, do not take the negative root.
Step 2
Why this answer is correct
The correct answer is B. (10). ((x+5)2 =225) and (x+5=15), so (x=10). If the positive root is given, do not take the negative root.
Step 3
Exam Tip
((x+5)2 =225) और (x+5=15), इसलिए (x=10) है। धनात्मक मूल दिया हो तो ऋणात्मक मूल नहीं लेना है।
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किसी संख्या में (3) जोड़ने पर प्राप्त संख्या का वर्ग (100) है। धनात्मक मूल के अनुसार संख्या क्या है?
When (3) is added to a number, the square of the result is (100). According to the positive root, what is the number?
#quadratic equations
#number problem
#square
A (5)
B (6)
C (7)
D (8)
Explanation opens after your attempt
Step 1
Concept
((x+3)2 =100) gives (x+3=10), so (x=7). If the question says positive root, take (10).
Step 2
Why this answer is correct
The correct answer is C. (7). ((x+3)2 =100) gives (x+3=10), so (x=7). If the question says positive root, take (10).
Step 3
Exam Tip
((x+3)2 =100) से (x+3=10), इसलिए (x=7) है। प्रश्न में धनात्मक मूल कहा हो तो (10) लें।
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यदि \(x^2-25x+q=0\) का एक मूल (10) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-25x+q=0\) is (10), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (15)
B (10)
C (25)
D (-15)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (25), so the other root is (25-10=15). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (15). The sum of roots is (25), so the other root is (25-10=15). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (25) है, इसलिए दूसरा मूल (25-10=15) होगा। परीक्षा में एक मूल दिया हो तो योग का प्रयोग करें।
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यदि \(x^2-23x+q=0\) का एक मूल (9) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-23x+q=0\) is (9), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (14)
B (9)
C (23)
D (-14)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (23), so the other root is (23-9=14). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (14). The sum of roots is (23), so the other root is (23-9=14). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (23) है, इसलिए दूसरा मूल (23-9=14) होगा। परीक्षा में एक मूल दिया हो तो योग का प्रयोग करें।
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यदि \(x^2-21x+q=0\) का एक मूल (8) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-21x+q=0\) is (8), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (13)
B (8)
C (21)
D (-13)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (21), so the other root is (21-8=13). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (13). The sum of roots is (21), so the other root is (21-8=13). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (21) है, इसलिए दूसरा मूल (21-8=13) होगा। परीक्षा में एक मूल दिया हो तो योग का उपयोग करें।
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यदि \(x^2-15x+q=0\) का एक मूल (6) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-15x+q=0\) is (6), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (9)
B (6)
C (15)
D (-9)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (15), so the other root is (15-6=9). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (9). The sum of roots is (15), so the other root is (15-6=9). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (15) है, इसलिए दूसरा मूल (15-6=9) होगा। परीक्षा में एक मूल दिया हो तो योग का प्रयोग करें।
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यदि \(x^2-9x+q=0\) का एक मूल (4) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-9x+q=0\) is (4), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (5)
B (4)
C (9)
D (-5)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (9), so the other root is (9-4=5). In exams, use the sum when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (5). The sum of roots is (9), so the other root is (9-4=5). In exams, use the sum when one root is given.
Step 3
Exam Tip
मूलों का योग (9) है, इसलिए दूसरा मूल (9-4=5) होगा। परीक्षा में एक मूल दिया हो तो योग का प्रयोग करें।
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यदि \(x^2-5x+q=0\) का एक मूल (2) है, तो दूसरा मूल क्या होगा?
If one root of \(x^2-5x+q=0\) is (2), what will be the other root?
#quadratic
#roots
#sum-of-roots
A (3)
B (2)
C (5)
D (-3)
Explanation opens after your attempt
Step 1
Concept
The sum of roots is (5), so the other root is (5-2=3). In exams, use sum or product when one root is given.
Step 2
Why this answer is correct
The correct answer is A. (3). The sum of roots is (5), so the other root is (5-2=3). In exams, use sum or product when one root is given.
Step 3
Exam Tip
मूलों का योग (5) है, इसलिए दूसरा मूल (5-2=3) होगा। परीक्षा में एक मूल दिया हो तो योग या गुणनफल का प्रयोग करें।
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यदि (x=0), \(ax^2+bx+c=0\) की जड़ है, तो कौन-सी शर्त निश्चित रूप से सही है?
If (x=0) is a root of \(ax^2+bx+c=0\), which condition must be true?
#quadratic-roots
#zero-root
#root-verification
A (c=0)
B (b=0)
C (a=0)
D (a+b=0)
Explanation opens after your attempt
Step 1
Concept
Putting (x=0) gives (c=0). Thus the direct condition for zero to be a root is (c=0).
Step 2
Why this answer is correct
The correct answer is A. (c=0). Putting (x=0) gives (c=0). Thus the direct condition for zero to be a root is (c=0).
Step 3
Exam Tip
(x=0) रखने पर समीकरण (c=0) बनता है। इसलिए शून्य जड़ होने की सीधी शर्त (c=0) है।
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यदि \(x^2+ax+12=0\) की एक जड़ दूसरी जड़ से (3) अधिक है, तो (a) के संभव मान क्या हैं?
If one root of \(x^2+ax+12=0\) is (3) more than the other root, what are the possible values of (a)?
#quadratic-roots
#difference-of-roots
#parameter
A \(\sqrt{57}\) और \(-\sqrt{57}\) / \(\sqrt{57}\) and \(-\sqrt{57}\)
B \(\sqrt{21}\) और \(-\sqrt{21}\) / \(\sqrt{21}\) and \(-\sqrt{21}\)
C (3) और (-3) / (3) and (-3)
D (12) और (-12) / (12) and (-12)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{57}\) और \(-\sqrt{57}\) / \(\sqrt{57}\) and \(-\sqrt{57}\)
Step 1
Concept
Let the roots be (r) and (r+3). Then (r(r+3)=12), giving the sum as \(\pm\sqrt{57}\), so \(a=\mp\sqrt{57}\).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{57}\) और \(-\sqrt{57}\) / \(\sqrt{57}\) and \(-\sqrt{57}\). Let the roots be (r) and (r+3). Then (r(r+3)=12), giving the sum as \(\pm\sqrt{57}\), so \(a=\mp\sqrt{57}\).
Step 3
Exam Tip
जड़ें (r) और (r+3) मानने पर (r(r+3)=12) मिलता है। इससे जड़ों का योग \(\pm\sqrt{57}\) होता है, इसलिए \(a=\mp\sqrt{57}\)।
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यदि (x-2 +(k-3)x+k=0) की एक जड़ दूसरी जड़ की दुगुनी है, तो (k) का मान क्या होगा?
If one root of (x-2 +(k-3)x+k=0) is twice the other root, what is the value of (k)?
#quadratic-roots
#roots-ratio
#parameter
A \(\frac{21+3\sqrt{33}}{4}\) या \(\frac{21-3\sqrt{33}}{4}\) / \(\frac{21+3\sqrt{33}}{4}\) or \(\frac{21-3\sqrt{33}}{4}\)
B \(\frac{3+\sqrt{33}}{4}\) या \(\frac{3-\sqrt{33}}{4}\) / \(\frac{3+\sqrt{33}}{4}\) or \(\frac{3-\sqrt{33}}{4}\)
C (6) या (3) / (6) or (3)
D (9) या (2) / (9) or (2)
Explanation opens after your attempt
Correct Answer
A. \(\frac{21+3\sqrt{33}}{4}\) या \(\frac{21-3\sqrt{33}}{4}\) / \(\frac{21+3\sqrt{33}}{4}\) or \(\frac{21-3\sqrt{33}}{4}\)
Step 1
Concept
Taking the roots as (r) and (2r), we get (3r=3-k) and \(2r^2=k\). Solving \(2k^2-21k+18=0\) gives the two listed values.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{21+3\sqrt{33}}{4}\) या \(\frac{21-3\sqrt{33}}{4}\) / \(\frac{21+3\sqrt{33}}{4}\) or \(\frac{21-3\sqrt{33}}{4}\). Taking the roots as (r) and (2r), we get (3r=3-k) and \(2r^2=k\). Solving \(2k^2-21k+18=0\) gives the two listed values.
Step 3
Exam Tip
जड़ें (r) और (2r) मानने पर (3r=3-k) और \(2r^2=k\) मिलता है। हल करने पर \(2k^2-21k+18=0\), इसलिए दिए गए दोनों मान मिलते हैं।
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यदि \(x^2-13x+42=0\) के मूलों में छोटा मूल \(\alpha\) और बड़ा मूल \(\beta\) है तो \(\beta-\alpha\) क्या है?
If the smaller root of \(x^2-13x+42=0\) is \(\alpha\) and the larger root is \(\beta\), what is \(\beta-\alpha\)?
#roots
#ordered_roots
#difference
A (1)
B (13)
C (42)
D (6)
Explanation opens after your attempt
Step 1
Concept
The roots are (6) and (7). Thus the smaller root is (6) and the larger root is (7), so \(\beta-\alpha=1\).
Step 2
Why this answer is correct
The correct answer is A. (1). The roots are (6) and (7). Thus the smaller root is (6) and the larger root is (7), so \(\beta-\alpha=1\).
Step 3
Exam Tip
समीकरण के मूल (6) और (7) हैं। इसलिए छोटा मूल (6) और बड़ा मूल (7) है तथा \(\beta-\alpha=1\) है।
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यदि \(x^2+ax+a=0\) का एक मूल (2) है तो दूसरा मूल क्या होगा?
If one root of \(x^2+ax+a=0\) is (2), what will be the other root?
#roots
#parameter
#other_root
A \(-\frac{2}{3}\)
B \(\frac{2}{3}\)
C (3)
D (-3)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{2}{3}\)
Step 1
Concept
Putting (x=2) gives (4+3a=0), so \(a=-\frac{4}{3}\). The product is (a), so the other root is \(-\frac{2}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{2}{3}\). Putting (x=2) gives (4+3a=0), so \(a=-\frac{4}{3}\). The product is (a), so the other root is \(-\frac{2}{3}\).
Step 3
Exam Tip
(x=2) रखने पर (4+3a=0) से \(a=-\frac{4}{3}\) है। गुणनफल (a) है इसलिए दूसरा मूल \(-\frac{2}{3}\) होगा।
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यदि \(x^2-11x+30=0\) के मूलों में छोटा मूल \(\alpha\) और बड़ा मूल \(\beta\) है तो \(\beta-\alpha\) क्या है?
If the smaller root of \(x^2-11x+30=0\) is \(\alpha\) and the larger root is \(\beta\), what is \(\beta-\alpha\)?
#roots
#ordered_roots
#difference
A (1)
B (11)
C (30)
D (5)
Explanation opens after your attempt
Step 1
Concept
The roots are (5) and (6). Thus the smaller root is (5) and the larger root is (6), so \(\beta-\alpha=1\).
Step 2
Why this answer is correct
The correct answer is A. (1). The roots are (5) and (6). Thus the smaller root is (5) and the larger root is (6), so \(\beta-\alpha=1\).
Step 3
Exam Tip
समीकरण के मूल (5) और (6) हैं। इसलिए छोटा मूल (5) और बड़ा मूल (6) है तथा \(\beta-\alpha=1\) है।
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यदि \(x^2+ax+a=0\) का एक मूल (1) है तो दूसरा मूल क्या होगा?
If one root of \(x^2+ax+a=0\) is (1), what will be the other root?
#roots
#parameter
#other_root
A \(-\frac{1}{2}\)
B \(\frac{1}{2}\)
C (2)
D (-2)
Explanation opens after your attempt
Correct Answer
A. \(-\frac{1}{2}\)
Step 1
Concept
Putting (x=1) gives (1+2a=0), so \(a=-\frac{1}{2}\). The product is (a), so the other root is \(-\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(-\frac{1}{2}\). Putting (x=1) gives (1+2a=0), so \(a=-\frac{1}{2}\). The product is (a), so the other root is \(-\frac{1}{2}\).
Step 3
Exam Tip
(x=1) रखने पर (1+2a=0) से \(a=-\frac{1}{2}\) है। गुणनफल (a) है इसलिए दूसरा मूल \(-\frac{1}{2}\) होगा।
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समीकरण \(x^2-17x+70=0\) का एक मूल (7) है तो दूसरा मूल क्या है?
If one root of \(x^2-17x+70=0\) is (7), what is the other root?
#roots
#other_root
#product
A (10)
B (7)
C (17)
D (70)
Explanation opens after your attempt
Step 1
Concept
The product of roots is (70) and one root is (7). The other root is \(\frac{70}{7}=10\).
Step 2
Why this answer is correct
The correct answer is A. (10). The product of roots is (70) and one root is (7). The other root is \(\frac{70}{7}=10\).
Step 3
Exam Tip
मूलों का गुणनफल (70) है और एक मूल (7) है। दूसरा मूल \(\frac{70}{7}=10\) होगा।
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समीकरण \(x^2-15x+54=0\) का एक मूल (6) है तो दूसरा मूल क्या है?
If one root of \(x^2-15x+54=0\) is (6), what is the other root?
#roots
#other_root
#product
A (9)
B (6)
C (15)
D (54)
Explanation opens after your attempt
Step 1
Concept
The product of roots is (54) and one root is (6). Therefore the other root is \(\frac{54}{6}=9\).
Step 2
Why this answer is correct
The correct answer is A. (9). The product of roots is (54) and one root is (6). Therefore the other root is \(\frac{54}{6}=9\).
Step 3
Exam Tip
मूलों का गुणनफल (54) है और एक मूल (6) है। इसलिए दूसरा मूल \(\frac{54}{6}=9\) है।
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समीकरण \(x^2-13x+40=0\) का एक मूल (5) है तो दूसरा मूल क्या है?
If one root of \(x^2-13x+40=0\) is (5), what is the other root?
#roots
#other_root
#product
A (8)
B (5)
C (13)
D (40)
Explanation opens after your attempt
Step 1
Concept
The product of roots is (40) and one root is (5). Therefore the other root is \(\frac{40}{5}=8\).
Step 2
Why this answer is correct
The correct answer is A. (8). The product of roots is (40) and one root is (5). Therefore the other root is \(\frac{40}{5}=8\).
Step 3
Exam Tip
मूलों का गुणनफल (40) है और एक मूल (5) है। इसलिए दूसरा मूल \(\frac{40}{5}=8\) है।
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यदि मूलों का योग (12) है और एक मूल (5) है तो दूसरा मूल क्या है?
If the sum of roots is (12) and one root is (5), what is the other root?
#roots
#other_root
#sum
A (5)
B (7)
C (12)
D (17)
Explanation opens after your attempt
Step 1
Concept
The other root is (12-5=7). Subtract the given root from the sum.
Step 2
Why this answer is correct
The correct answer is B. (7). The other root is (12-5=7). Subtract the given root from the sum.
Step 3
Exam Tip
दूसरा मूल (12-5=7) है। योग में से दिया हुआ मूल घटाएं।
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यदि एक मूल (6) है और मूलों का गुणनफल (48) है तो दूसरा मूल क्या होगा?
If one root is (6) and the product of roots is (48), what is the other root?
#roots
#other_root
#product
A (6)
B (8)
C (42)
D (48)
Explanation opens after your attempt
Step 1
Concept
The other root is \(\frac{48}{6}=8\). Divide the product by the given root.
Step 2
Why this answer is correct
The correct answer is B. (8). The other root is \(\frac{48}{6}=8\). Divide the product by the given root.
Step 3
Exam Tip
दूसरा मूल \(\frac{48}{6}=8\) होगा। गुणनफल को दिए हुए मूल से भाग करें।
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समीकरण \(x^2-11x+30=0\) का एक मूल (5) है तो दूसरा मूल क्या है?
If one root of \(x^2-11x+30=0\) is (5), what is the other root?
#roots
#other_root
#factorisation
A (5)
B (6)
C (11)
D (30)
Explanation opens after your attempt
Step 1
Concept
(x-2 -11x+30=(x-5)(x-6)). Therefore the other root is (6).
Step 2
Why this answer is correct
The correct answer is B. (6). (x-2 -11x+30=(x-5)(x-6)). Therefore the other root is (6).
Step 3
Exam Tip
(x-2 -11x+30=(x-5)(x-6)) है। इसलिए दूसरा मूल (6) है।
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यदि मूलों का योग (8) है और एक मूल (3) है तो दूसरा मूल क्या है?
If the sum of roots is (8) and one root is (3), what is the other root?
#roots
#other_root
#sum
A (3)
B (5)
C (8)
D (11)
Explanation opens after your attempt
Step 1
Concept
The other root is (8-3=5). Subtract the given root from the sum.
Step 2
Why this answer is correct
The correct answer is B. (5). The other root is (8-3=5). Subtract the given root from the sum.
Step 3
Exam Tip
दूसरा मूल (8-3=5) है। योग में से दिया हुआ मूल घटाएं।
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यदि एक मूल (5) है और मूलों का गुणनफल (35) है तो दूसरा मूल क्या होगा?
If one root is (5) and the product of roots is (35), what is the other root?
#roots
#other_root
#product
A (5)
B (7)
C (30)
D (35)
Explanation opens after your attempt
Step 1
Concept
The other root is \(\frac{35}{5}=7\). Divide the product by the given root.
Step 2
Why this answer is correct
The correct answer is B. (7). The other root is \(\frac{35}{5}=7\). Divide the product by the given root.
Step 3
Exam Tip
दूसरा मूल \(\frac{35}{5}=7\) होगा। गुणनफल में दिए हुए मूल से भाग करें।
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समीकरण \(x^2-9x+18=0\) का एक मूल (3) है तो दूसरा मूल क्या है?
If one root of \(x^2-9x+18=0\) is (3), what is the other root?
#roots
#other_root
#factorisation
A (3)
B (6)
C (9)
D (18)
Explanation opens after your attempt
Step 1
Concept
(x-2 -9x+18=(x-3)(x-6)). Therefore the other root is (6).
Step 2
Why this answer is correct
The correct answer is B. (6). (x-2 -9x+18=(x-3)(x-6)). Therefore the other root is (6).
Step 3
Exam Tip
(x-2 -9x+18=(x-3)(x-6)) है। इसलिए दूसरा मूल (6) है।
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यदि मूलों का योग (5) है और एक मूल (2) है तो दूसरा मूल क्या है?
If the sum of roots is (5) and one root is (2) then what is the other root?
#roots
#other_root
#sum
A (2)
B (3)
C (5)
D (7)
Explanation opens after your attempt
Step 1
Concept
The other root is (5-2=3). Subtract the given root from the sum.
Step 2
Why this answer is correct
The correct answer is B. (3). The other root is (5-2=3). Subtract the given root from the sum.
Step 3
Exam Tip
दूसरा मूल (5-2=3) है। योग में से दिए हुए मूल को घटाएं।
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यदि एक मूल (3) है और मूलों का गुणनफल (12) है तो दूसरा मूल क्या होगा?
If one root is (3) and the product of roots is (12) then what is the other root?
#roots
#other_root
#product
A (3)
B (4)
C (9)
D (12)
Explanation opens after your attempt
Step 1
Concept
The other root is \(\frac{12}{3}=4\). In product questions divide by the given root.
Step 2
Why this answer is correct
The correct answer is B. (4). The other root is \(\frac{12}{3}=4\). In product questions divide by the given root.
Step 3
Exam Tip
दूसरा मूल \(\frac{12}{3}=4\) होगा। गुणनफल वाले प्रश्न में दिए मूल से भाग दें।
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समीकरण \(x^2-6x+8=0\) का एक मूल (4) है तो दूसरा मूल क्या है?
If one root of \(x^2-6x+8=0\) is (4) then what is the other root?
#roots
#other_root
#factorisation
A (1)
B (2)
C (-2)
D (8)
Explanation opens after your attempt
Step 1
Concept
(x-2 -6x+8=(x-4)(x-2)) so the other root is (2). Use the given root to find the other factor.
Step 2
Why this answer is correct
The correct answer is B. (2). (x-2 -6x+8=(x-4)(x-2)) so the other root is (2). Use the given root to find the other factor.
Step 3
Exam Tip
(x-2 -6x+8=(x-4)(x-2)) इसलिए दूसरा मूल (2) है। दिए गए एक मूल से दूसरा गुणनखंड खोजें।
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किस समीकरण में (x=-2) मूल नहीं है?
In which equation is (x=-2) not a root?
#quadratic-equations
#root-check
#not-root
#medium
A \(x^2+2x=0\)
B \(x^2+5x+6=0\)
C \(2x^2+3x-2=0\)
D \(x^2-2x+4=0\)
Explanation opens after your attempt
Correct Answer
D. \(x^2-2x+4=0\)
Step 1
Concept
Putting (x=-2) gives \(4+4+4=12\neq0\). To check a non-root, use substitution too.
Step 2
Why this answer is correct
The correct answer is D. \(x^2-2x+4=0\). Putting (x=-2) gives \(4+4+4=12\neq0\). To check a non-root, use substitution too.
Step 3
Exam Tip
(x=-2) रखने पर \(4+4+4=12\neq0\) मिलता है। मूल न होने की जांच भी प्रतिस्थापन से करें।
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किस समीकरण में (x=1) मूल नहीं है?
In which equation is (x=1) not a root?
#quadratic-equations
#root-check
#not-root
#medium
A \(x^2-1=0\)
B \(x^2-3x+2=0\)
C \(2x^2+x-3=0\)
D \(x^2+x+1=0\)
Explanation opens after your attempt
Correct Answer
D. \(x^2+x+1=0\)
Step 1
Concept
Putting (x=1) gives \(1+1+1=3\neq 0\). To check when a value is not a root, use substitution too.
Step 2
Why this answer is correct
The correct answer is D. \(x^2+x+1=0\). Putting (x=1) gives \(1+1+1=3\neq 0\). To check when a value is not a root, use substitution too.
Step 3
Exam Tip
(x=1) रखने पर \(1+1+1=3\neq 0\) मिलता है। किसी विकल्प में मूल न होने की जांच भी प्रतिस्थापन से करें।
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\(\sqrt{13}\) का वर्ग किसके बराबर है?
The square of \(\sqrt{13}\) is equal to what?
#real-numbers
#square-root
#square
A (169)
B \(\sqrt{13}\)
C (13)
D (26)
Explanation opens after your attempt
Step 1
Concept
Squaring a square root gives the number inside it.
Step 2
Why this answer is correct
(\(\sqrt{13}\)2 =13).
Step 3
Exam Tip
Apply (\(\sqrt{a}\)2 =a) directly. चरण 1: वर्गमूल का वर्ग करने पर अंदर की संख्या मिलती है। चरण 2: (\(\sqrt{13}\)2 =13)। चरण 3: (\(\sqrt{a}\)2 =a) को सीधे लागू करें।
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