100 results found for "square root" in Class 10.
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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\(7x^2=175\) 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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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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समीकरण \(9x^2-6x+1=0\) में समान मूल का मान क्या है?
What is the equal root in \(9x^2-6x+1=0\)?
#quadratic-equations
#equal-root-value
#perfect-square
A \(x=\frac{1}{3}\)
B \(x=-\frac{1}{3}\)
C (x=3)
D (x=1)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{1}{3}\)
Step 1
Concept
Here (9x-2 -6x+1=(3x-1)2 ). Therefore the equal root is \(x=\frac{1}{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{1}{3}\). Here (9x-2 -6x+1=(3x-1)2 ). Therefore the equal root is \(x=\frac{1}{3}\).
Step 3
Exam Tip
यहाँ (9x-2 -6x+1=(3x-1)2 ) है। इसलिए समान मूल \(x=\frac{1}{3}\) है।
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समीकरण \(4x^2-20x+25=0\) में समान मूल का मान क्या है?
What is the equal root in \(4x^2-20x+25=0\)?
#quadratic-equations
#equal-root-value
#perfect-square
A \(x=\frac{5}{2}\)
B \(x=-\frac{5}{2}\)
C (x=5)
D \(x=\frac{1}{2}\)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{5}{2}\)
Step 1
Concept
Here (D=(-20)2 -4(4)(25)=0), and ((2x-5)2 =0). So the equal root is \(x=\frac{5}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{5}{2}\). Here (D=(-20)2 -4(4)(25)=0), and ((2x-5)2 =0). So the equal root is \(x=\frac{5}{2}\).
Step 3
Exam Tip
यहाँ (D=(-20)2 -4(4)(25)=0) और ((2x-5)2 =0) है। इसलिए समान मूल \(x=\frac{5}{2}\) है।
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\(49x^2-42x+9=0\) का मूल क्या है?
What is the root of \(49x^2-42x+9=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{3}{7}\)
B \(x=-\frac{3}{7}\)
C \(x=\frac{7}{3}\)
D (x=3)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{7}\)
Step 1
Concept
((7x-3)2 =0), so (7x-3=0) and \(x=\frac{3}{7}\). In exams, solve the linear equation after square form.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{7}\). ((7x-3)2 =0), so (7x-3=0) and \(x=\frac{3}{7}\). In exams, solve the linear equation after square form.
Step 3
Exam Tip
((7x-3)2 =0), इसलिए (7x-3=0) और \(x=\frac{3}{7}\) है। परीक्षा में वर्ग रूप के बाद रैखिक समीकरण हल करें।
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\(36x^2-60x+25=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(36x^2-60x+25=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{5}{6}\)
B \(x=-\frac{5}{6}\)
C \(x=\frac{6}{5}\)
D (x=5)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{5}{6}\)
Step 1
Concept
((6x-5)2 =0), so (6x-5=0) and \(x=\frac{5}{6}\). In exams, write the repeated root as a correct fraction.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{5}{6}\). ((6x-5)2 =0), so (6x-5=0) and \(x=\frac{5}{6}\). In exams, write the repeated root as a correct fraction.
Step 3
Exam Tip
((6x-5)2 =0), इसलिए (6x-5=0) और \(x=\frac{5}{6}\) है। परीक्षा में दोहराए हुए मूल को सही भिन्न में लिखें।
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\(25x^2-20x+4=0\) का मूल क्या है?
What is the root of \(25x^2-20x+4=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{2}{5}\)
B \(x=-\frac{2}{5}\)
C \(x=\frac{5}{2}\)
D (x=2)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{2}{5}\)
Step 1
Concept
((5x-2)2 =0), so (5x-2=0) and \(x=\frac{2}{5}\). In exams, solve the linear equation after square form.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{2}{5}\). ((5x-2)2 =0), so (5x-2=0) and \(x=\frac{2}{5}\). In exams, solve the linear equation after square form.
Step 3
Exam Tip
((5x-2)2 =0), इसलिए (5x-2=0) और \(x=\frac{2}{5}\) है। परीक्षा में वर्ग रूप के बाद रैखिक समीकरण हल करें।
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\(16x^2-24x+9=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(16x^2-24x+9=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{3}{4}\)
B \(x=-\frac{3}{4}\)
C \(x=\frac{4}{3}\)
D (x=3)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{4}\)
Step 1
Concept
((4x-3)2 =0), so (4x-3=0) and \(x=\frac{3}{4}\). In exams, write the repeated root as a correct fraction.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{4}\). ((4x-3)2 =0), so (4x-3=0) and \(x=\frac{3}{4}\). In exams, write the repeated root as a correct fraction.
Step 3
Exam Tip
((4x-3)2 =0), इसलिए (4x-3=0) और \(x=\frac{3}{4}\) है। परीक्षा में दोहराए हुए मूल को भी सही भिन्न में लिखें।
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\(9x^2-30x+25=0\) का मूल क्या है?
What is the root of \(9x^2-30x+25=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{5}{3}\)
B \(x=-\frac{5}{3}\)
C \(x=\frac{3}{5}\)
D (x=5)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{5}{3}\)
Step 1
Concept
((3x-5)2 =0), so (3x-5=0) and \(x=\frac{5}{3}\). In exams, solve the linear equation after square form.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{5}{3}\). ((3x-5)2 =0), so (3x-5=0) and \(x=\frac{5}{3}\). In exams, solve the linear equation after square form.
Step 3
Exam Tip
((3x-5)2 =0), इसलिए (3x-5=0) और \(x=\frac{5}{3}\) है। परीक्षा में वर्ग रूप के बाद रैखिक समीकरण हल करें।
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\(4x^2-12x+9=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(4x^2-12x+9=0\)?
#quadratic
#repeated-root
#perfect-square
A \(x=\frac{3}{2}\)
B \(x=-\frac{3}{2}\)
C (x=3)
D \(x=\frac{2}{3}\)
Explanation opens after your attempt
Correct Answer
A. \(x=\frac{3}{2}\)
Step 1
Concept
((2x-3)2 =0), so (2x-3=0) and \(x=\frac{3}{2}\). In exams, write the repeated root with the correct value.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{2}\). ((2x-3)2 =0), so (2x-3=0) and \(x=\frac{3}{2}\). In exams, write the repeated root with the correct value.
Step 3
Exam Tip
((2x-3)2 =0), इसलिए (2x-3=0) और \(x=\frac{3}{2}\) है। परीक्षा में दोहराए हुए मूल को भी सही मान से लिखें।
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\(x^2+18x+81=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(x^2+18x+81=0\)?
#quadratic
#repeated-root
#perfect-square
A (x=-9)
B (x=9)
C (x=-18)
D (x=18)
Explanation opens after your attempt
Step 1
Concept
((x+9)2 =0), so the repeated root is (-9). In exams, a perfect square equation has equal roots.
Step 2
Why this answer is correct
The correct answer is A. (x=-9). ((x+9)2 =0), so the repeated root is (-9). In exams, a perfect square equation has equal roots.
Step 3
Exam Tip
((x+9)2 =0), इसलिए दोहराया हुआ मूल (-9) है। परीक्षा में पूर्ण वर्ग समीकरण में दोनों मूल समान होते हैं।
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\(x^2+14x+49=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(x^2+14x+49=0\)?
#quadratic
#repeated-root
#perfect-square
A (x=-7)
B (x=7)
C (x=-14)
D (x=14)
Explanation opens after your attempt
Step 1
Concept
((x+7)2 =0), so the repeated root is (-7). In exams, ((x+a)2 =0) gives (x=-a).
Step 2
Why this answer is correct
The correct answer is A. (x=-7). ((x+7)2 =0), so the repeated root is (-7). In exams, ((x+a)2 =0) gives (x=-a).
Step 3
Exam Tip
((x+7)2 =0), इसलिए दोहराया हुआ मूल (-7) है। परीक्षा में ((x+a)2 =0) से (x=-a) मिलता है।
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\(x^2-12x+36=0\) का मूल क्या होगा?
What will be the root of \(x^2-12x+36=0\)?
#quadratic
#repeated-root
#perfect-square
A (x=6)
B (x=-6)
C (x=12)
D (x=-12)
Explanation opens after your attempt
Step 1
Concept
((x-6)2 =0), so both equal roots are (x=6). In exams, a perfect square equation gives a repeated root.
Step 2
Why this answer is correct
The correct answer is A. (x=6). ((x-6)2 =0), so both equal roots are (x=6). In exams, a perfect square equation gives a repeated root.
Step 3
Exam Tip
((x-6)2 =0), इसलिए दोनों समान मूल (x=6) हैं। परीक्षा में पूर्ण वर्ग समीकरण में दोहराया हुआ मूल मिलता है।
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\(x^2-6x+9=0\) का दोहराया हुआ मूल क्या है?
What is the repeated root of \(x^2-6x+9=0\)?
#quadratic
#repeated-root
#perfect-square
A (x=3)
B (x=-3)
C (x=6)
D (x=9)
Explanation opens after your attempt
Step 1
Concept
(x-2 -6x+9=(x-3)2 ), so the repeated root is (3). In exams, a perfect square gives equal roots.
Step 2
Why this answer is correct
The correct answer is A. (x=3). (x-2 -6x+9=(x-3)2 ), so the repeated root is (3). In exams, a perfect square gives equal roots.
Step 3
Exam Tip
(x-2 -6x+9=(x-3)2 ), इसलिए दोहराया हुआ मूल (3) है। परीक्षा में पूर्ण वर्ग में दोनों मूल समान होते हैं।
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एक वर्गाकार मैदान की भुजा (4) मीटर बढ़ाने पर क्षेत्रफल (96) वर्ग मीटर बढ़ जाता है। मूल भुजा कितनी थी?
When the side of a square field is increased by (4) m, its area increases by (96) square m. What was the original side?
#quadratic equations
#square
#area increase
A (8) मीटर / (8) m
B (10) मीटर / (10) m
C (12) मीटर / (12) m
D (14) मीटर / (14) m
Explanation opens after your attempt
Correct Answer
B. (10) मीटर / (10) m
Step 1
Concept
If the original side is (x), then ((x+4)2 -x-2 =96). Thus (8x+16=96), giving (x=10).
Step 2
Why this answer is correct
The correct answer is B. (10) मीटर / (10) m. If the original side is (x), then ((x+4)2 -x-2 =96). Thus (8x+16=96), giving (x=10).
Step 3
Exam Tip
मूल भुजा (x) हो तो ((x+4)2 -x-2 =96)। इससे (8x+16=96) और (x=10) मिलता है।
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एक वर्गाकार मैदान के अंदर समान चौड़ाई (x) मीटर की सीमा छोड़ने के बाद अंदर का वर्ग (60) मीटर भुजा का बचता है। बाहरी भुजा (92) मीटर है। (x) क्या है?
Inside a square field, after leaving a uniform border of width (x) m, the inner square has side (60) m. The outer side is (92) m. What is (x)?
#quadratic equations
#square border
#application
A (12) मीटर / (12) m
B (14) मीटर / (14) m
C (16) मीटर / (16) m
D (18) मीटर / (18) m
Explanation opens after your attempt
Correct Answer
C. (16) मीटर / (16) m
Step 1
Concept
The inner side is (92-2x) and (92-2x=60), so (x=16). The border is subtracted from both sides.
Step 2
Why this answer is correct
The correct answer is C. (16) मीटर / (16) m. The inner side is (92-2x) and (92-2x=60), so (x=16). The border is subtracted from both sides.
Step 3
Exam Tip
अंदर की भुजा (92-2x) है और (92-2x=60), इसलिए (x=16) है। सीमा दोनों तरफ से घटती है।
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एक तार को वर्ग के रूप में मोड़ने पर क्षेत्रफल (3025) वर्ग सेमी है। तार की लंबाई क्या है?
A wire is bent into a square and the area is (3025) square cm. What is the length of the wire?
#quadratic equations
#square
#wire
A (200) सेमी / (200) cm
B (210) सेमी / (210) cm
C (220) सेमी / (220) cm
D (240) सेमी / (240) cm
Explanation opens after your attempt
Correct Answer
C. (220) सेमी / (220) cm
Step 1
Concept
If the side of the square is (x), then \(x^2=3025\), so (x=55). The wire length is the perimeter (4x=220) cm.
Step 2
Why this answer is correct
The correct answer is C. (220) सेमी / (220) cm. If the side of the square is (x), then \(x^2=3025\), so (x=55). The wire length is the perimeter (4x=220) cm.
Step 3
Exam Tip
वर्ग की भुजा (x) हो तो \(x^2=3025\), इसलिए (x=55) है। तार की लंबाई परिमाप (4x=220) सेमी होगी।
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एक वर्ग की भुजा (x) सेमी है। भुजा (9) सेमी घटाने पर क्षेत्रफल (1681) वर्ग सेमी हो जाता है। मूल भुजा क्या है?
The side of a square is (x) cm. When the side is reduced by (9) cm, the area becomes (1681) square cm. What is the original side?
#quadratic equations
#square
#area change
A (41) सेमी / (41) cm
B (45) सेमी / (45) cm
C (50) सेमी / (50) cm
D (57) सेमी / (57) cm
Explanation opens after your attempt
Correct Answer
C. (50) सेमी / (50) cm
Step 1
Concept
((x-9)2 =1681) gives (x-9=41), so (x=50). Write the reduced side as (x-9).
Step 2
Why this answer is correct
The correct answer is C. (50) सेमी / (50) cm. ((x-9)2 =1681) gives (x-9=41), so (x=50). Write the reduced side as (x-9).
Step 3
Exam Tip
((x-9)2 =1681) से (x-9=41), इसलिए (x=50) है। घटाई गई भुजा को (x-9) लिखें।
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एक वर्गाकार टाइल की भुजा (x-12) सेमी है और क्षेत्रफल (1849) वर्ग सेमी है। (x) क्या है?
The side of a square tile is (x-12) cm and its area is (1849) square cm. What is (x)?
#quadratic equations
#square
#tile
A (43)
B (49)
C (55)
D (60)
Explanation opens after your attempt
Step 1
Concept
((x-12)2 =1849) and (x-12=43), so (x=55). Remember that the side must be positive.
Step 2
Why this answer is correct
The correct answer is C. (55). ((x-12)2 =1849) and (x-12=43), so (x=55). Remember that the side must be positive.
Step 3
Exam Tip
((x-12)2 =1849) और (x-12=43), इसलिए (x=55) है। भुजा धनात्मक होने की शर्त याद रखें।
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एक वर्ग की भुजा (x+17) सेमी है। उसका क्षेत्रफल (3249) वर्ग सेमी है। (x) का मान क्या है?
The side of a square is (x+17) cm. Its area is (3249) square cm. What is the value of (x)?
#quadratic equations
#square
#area
A (34)
B (38)
C (40)
D (57)
Explanation opens after your attempt
Step 1
Concept
((x+17)2 =3249) gives (x+17=57), so (x=40). Take the positive square root for length.
Step 2
Why this answer is correct
The correct answer is C. (40). ((x+17)2 =3249) gives (x+17=57), so (x=40). Take the positive square root for length.
Step 3
Exam Tip
((x+17)2 =3249) से (x+17=57), इसलिए (x=40) है। लंबाई के लिए धनात्मक वर्गमूल लें।
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एक वर्गाकार मैदान के अंदर समान चौड़ाई (x) मीटर की सीमा छोड़ने के बाद अंदर का वर्ग (44) मीटर भुजा का बचता है। बाहरी भुजा (68) मीटर है। (x) क्या है?
Inside a square field, after leaving a uniform border of width (x) m, the inner square has side (44) m. The outer side is (68) m. What is (x)?
#quadratic equations
#square border
#application
A (8) मीटर / (8) m
B (10) मीटर / (10) m
C (12) मीटर / (12) m
D (14) मीटर / (14) m
Explanation opens after your attempt
Correct Answer
C. (12) मीटर / (12) m
Step 1
Concept
The inner side is (68-2x) and (68-2x=44), so (x=12). The border is subtracted from both sides.
Step 2
Why this answer is correct
The correct answer is C. (12) मीटर / (12) m. The inner side is (68-2x) and (68-2x=44), so (x=12). The border is subtracted from both sides.
Step 3
Exam Tip
अंदर की भुजा (68-2x) है और (68-2x=44), इसलिए (x=12) है। सीमा दोनों तरफ से घटती है।
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एक तार को वर्ग के रूप में मोड़ने पर क्षेत्रफल (2025) वर्ग सेमी है। तार की लंबाई क्या है?
A wire is bent into a square and the area is (2025) square cm. What is the length of the wire?
#quadratic equations
#square
#wire
A (160) सेमी / (160) cm
B (180) सेमी / (180) cm
C (200) सेमी / (200) cm
D (225) सेमी / (225) cm
Explanation opens after your attempt
Correct Answer
B. (180) सेमी / (180) cm
Step 1
Concept
If the side of the square is (x), then \(x^2=2025\), so (x=45). The wire length is the perimeter (4x=180) cm.
Step 2
Why this answer is correct
The correct answer is B. (180) सेमी / (180) cm. If the side of the square is (x), then \(x^2=2025\), so (x=45). The wire length is the perimeter (4x=180) cm.
Step 3
Exam Tip
वर्ग की भुजा (x) हो तो \(x^2=2025\), इसलिए (x=45) है। तार की लंबाई परिमाप (4x=180) सेमी होगी।
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एक वर्ग की भुजा (x) सेमी है। भुजा (7) सेमी घटाने पर क्षेत्रफल (1296) वर्ग सेमी हो जाता है। मूल भुजा क्या है?
The side of a square is (x) cm. When the side is reduced by (7) cm, the area becomes (1296) square cm. What is the original side?
#quadratic equations
#square
#area change
A (36) सेमी / (36) cm
B (40) सेमी / (40) cm
C (43) सेमी / (43) cm
D (49) सेमी / (49) cm
Explanation opens after your attempt
Correct Answer
C. (43) सेमी / (43) cm
Step 1
Concept
((x-7)2 =1296) gives (x-7=36), so (x=43). Write the reduced side as (x-7).
Step 2
Why this answer is correct
The correct answer is C. (43) सेमी / (43) cm. ((x-7)2 =1296) gives (x-7=36), so (x=43). Write the reduced side as (x-7).
Step 3
Exam Tip
((x-7)2 =1296) से (x-7=36), इसलिए (x=43) है। घटाई गई भुजा को (x-7) लिखें।
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एक वर्गाकार टाइल की भुजा (x-13) सेमी है और क्षेत्रफल (1600) वर्ग सेमी है। (x) क्या है?
The side of a square tile is (x-13) cm and its area is (1600) square cm. What is (x)?
#quadratic equations
#square
#tile
A (40)
B (47)
C (53)
D (60)
Explanation opens after your attempt
Step 1
Concept
((x-13)2 =1600) and (x-13=40), so (x=53). Remember that the side must be positive.
Step 2
Why this answer is correct
The correct answer is C. (53). ((x-13)2 =1600) and (x-13=40), so (x=53). Remember that the side must be positive.
Step 3
Exam Tip
((x-13)2 =1600) और (x-13=40), इसलिए (x=53) है। भुजा धनात्मक होने की शर्त याद रखें।
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एक वर्ग की भुजा (x+12) सेमी है। उसका क्षेत्रफल (1764) वर्ग सेमी है। (x) का मान क्या है?
The side of a square is (x+12) cm. Its area is (1764) square cm. What is the value of (x)?
#quadratic equations
#square
#area
A (24)
B (28)
C (30)
D (42)
Explanation opens after your attempt
Step 1
Concept
((x+12)2 =1764) gives (x+12=42), so (x=30). Take the positive square root for length.
Step 2
Why this answer is correct
The correct answer is C. (30). ((x+12)2 =1764) gives (x+12=42), so (x=30). Take the positive square root for length.
Step 3
Exam Tip
((x+12)2 =1764) से (x+12=42), इसलिए (x=30) है। लंबाई के लिए धनात्मक वर्गमूल लें।
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एक वर्गाकार मैदान के अंदर समान चौड़ाई (x) मीटर की सीमा छोड़ने के बाद अंदर का वर्ग (36) मीटर भुजा का बचता है। बाहरी भुजा (52) मीटर है। (x) क्या है?
Inside a square field, after leaving a uniform border of width (x) m, the inner square has side (36) m. The outer side is (52) m. What is (x)?
#quadratic equations
#square border
#application
A (6) मीटर / (6) m
B (8) मीटर / (8) m
C (10) मीटर / (10) m
D (12) मीटर / (12) m
Explanation opens after your attempt
Correct Answer
B. (8) मीटर / (8) m
Step 1
Concept
The inner side is (52-2x) and (52-2x=36), so (x=8). The border is subtracted from both sides.
Step 2
Why this answer is correct
The correct answer is B. (8) मीटर / (8) m. The inner side is (52-2x) and (52-2x=36), so (x=8). The border is subtracted from both sides.
Step 3
Exam Tip
अंदर की भुजा (52-2x) है और (52-2x=36), इसलिए (x=8) है। सीमा दोनों तरफ से घटती है।
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एक तार को वर्ग के रूप में मोड़ने पर क्षेत्रफल (1296) वर्ग सेमी है। तार की लंबाई क्या है?
A wire is bent into a square and the area is (1296) square cm. What is the length of the wire?
#quadratic equations
#square
#wire
A (120) सेमी / (120) cm
B (132) सेमी / (132) cm
C (144) सेमी / (144) cm
D (156) सेमी / (156) cm
Explanation opens after your attempt
Correct Answer
C. (144) सेमी / (144) cm
Step 1
Concept
If the side of the square is (x), then \(x^2=1296\), so (x=36). The wire length is the perimeter (4x=144) cm.
Step 2
Why this answer is correct
The correct answer is C. (144) सेमी / (144) cm. If the side of the square is (x), then \(x^2=1296\), so (x=36). The wire length is the perimeter (4x=144) cm.
Step 3
Exam Tip
वर्ग की भुजा (x) हो तो \(x^2=1296\), इसलिए (x=36) है। तार की लंबाई परिमाप (4x=144) सेमी होगी।
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एक वर्ग की भुजा (x) सेमी है। भुजा (5) सेमी घटाने पर क्षेत्रफल (784) वर्ग सेमी हो जाता है। मूल भुजा क्या है?
The side of a square is (x) cm. When the side is reduced by (5) cm, the area becomes (784) square cm. What is the original side?
#quadratic equations
#square
#area change
A (28) सेमी / (28) cm
B (30) सेमी / (30) cm
C (33) सेमी / (33) cm
D (35) सेमी / (35) cm
Explanation opens after your attempt
Correct Answer
C. (33) सेमी / (33) cm
Step 1
Concept
((x-5)2 =784) gives (x-5=28), so (x=33). Write the reduced side as (x-5).
Step 2
Why this answer is correct
The correct answer is C. (33) सेमी / (33) cm. ((x-5)2 =784) gives (x-5=28), so (x=33). Write the reduced side as (x-5).
Step 3
Exam Tip
((x-5)2 =784) से (x-5=28), इसलिए (x=33) है। घटाई गई भुजा को (x-5) लिखें।
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एक वर्गाकार टाइल की भुजा (x-8) सेमी है और क्षेत्रफल (900) वर्ग सेमी है। (x) क्या है?
The side of a square tile is (x-8) cm and its area is (900) square cm. What is (x)?
#quadratic equations
#square
#tile
A (30)
B (34)
C (38)
D (42)
Explanation opens after your attempt
Step 1
Concept
((x-8)2 =900) and (x-8=30), so (x=38). Remember that the side must be positive.
Step 2
Why this answer is correct
The correct answer is C. (38). ((x-8)2 =900) and (x-8=30), so (x=38). Remember that the side must be positive.
Step 3
Exam Tip
((x-8)2 =900) और (x-8=30), इसलिए (x=38) है। भुजा धनात्मक होने की शर्त याद रखें।
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एक वर्ग की भुजा (x+6) सेमी है। उसका क्षेत्रफल (1225) वर्ग सेमी है। (x) का मान क्या है?
The side of a square is (x+6) cm. Its area is (1225) square cm. What is the value of (x)?
#quadratic equations
#square
#area
A (23)
B (25)
C (29)
D (35)
Explanation opens after your attempt
Step 1
Concept
((x+6)2 =1225) gives (x+6=35), so (x=29). Take the positive square root for length.
Step 2
Why this answer is correct
The correct answer is C. (29). ((x+6)2 =1225) gives (x+6=35), so (x=29). Take the positive square root for length.
Step 3
Exam Tip
((x+6)2 =1225) से (x+6=35), इसलिए (x=29) है। लंबाई के लिए धनात्मक वर्गमूल लें।
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एक वर्ग का क्षेत्रफल उसके परिमाप से (96) अधिक है। वर्ग की भुजा क्या है?
The area of a square is (96) more than its perimeter. What is the side of the square?
#quadratic-equations
#word-problems
#square
A (12)
B (8)
C (16)
D (24)
Explanation opens after your attempt
Step 1
Concept
If the side is (x), then \(x^2=4x+96\). This gives \(x^2-4x-96=0\), so the positive side is (12).
Step 2
Why this answer is correct
The correct answer is A. (12). If the side is (x), then \(x^2=4x+96\). This gives \(x^2-4x-96=0\), so the positive side is (12).
Step 3
Exam Tip
यदि भुजा (x) है, तो \(x^2=4x+96\)। इससे \(x^2-4x-96=0\), इसलिए धनात्मक भुजा (12) है।
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एक वर्गाकार मैदान के अंदर समान चौड़ाई (x) मीटर की सीमा छोड़ने के बाद अंदर का वर्ग (20) मीटर भुजा का बचता है। बाहरी भुजा (28) मीटर है। (x) क्या है?
Inside a square field, after leaving a uniform border of width (x) m, the inner square has side (20) m. The outer side is (28) m. What is (x)?
#quadratic equations
#square border
#application
A (3) मीटर / (3) m
B (4) मीटर / (4) m
C (5) मीटर / (5) m
D (6) मीटर / (6) m
Explanation opens after your attempt
Correct Answer
B. (4) मीटर / (4) m
Step 1
Concept
The inner side is (28-2x) and (28-2x=20), so (x=4). The border is subtracted from both sides.
Step 2
Why this answer is correct
The correct answer is B. (4) मीटर / (4) m. The inner side is (28-2x) and (28-2x=20), so (x=4). The border is subtracted from both sides.
Step 3
Exam Tip
अंदर की भुजा (28-2x) है और (28-2x=20), इसलिए (x=4) है। सीमा दोनों तरफ से घटती है।
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एक तार को वर्ग के रूप में मोड़ने पर क्षेत्रफल (625) वर्ग सेमी है। तार की लंबाई क्या है?
A wire is bent into a square and the area is (625) square cm. What is the length of the wire?
#quadratic equations
#square
#wire
A (80) सेमी / (80) cm
B (90) सेमी / (90) cm
C (100) सेमी / (100) cm
D (125) सेमी / (125) cm
Explanation opens after your attempt
Correct Answer
C. (100) सेमी / (100) cm
Step 1
Concept
If the side of the square is (x), then \(x^2=625\), so (x=25). The wire length is the perimeter (4x=100) cm.
Step 2
Why this answer is correct
The correct answer is C. (100) सेमी / (100) cm. If the side of the square is (x), then \(x^2=625\), so (x=25). The wire length is the perimeter (4x=100) cm.
Step 3
Exam Tip
वर्ग की भुजा (x) हो तो \(x^2=625\), इसलिए (x=25) है। तार की लंबाई परिमाप (4x=100) सेमी होगी।
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