Taking \(10^4\) common in the numerator gives \(\dfrac{10^4(10-1)}{9\times 10^3}=10\). In exams, taking a common factor makes calculation easier.
Step 2
Why this answer is correct
The correct answer is A. (,10,). Taking \(10^4\) common in the numerator gives \(\dfrac{10^4(10-1)}{9\times 10^3}=10\). In exams, taking a common factor makes calculation easier.
Step 3
Exam Tip
ऊपर \(10^4\) common लेने पर \(\dfrac{10^4(10-1)}{9\times 10^3}=10\) मिलता है। परीक्षा में common factor लेने से गणना आसान होती है।
Taking \(7^4\) common in the numerator gives (\dfrac{74(7-1)}{74}=6). In exams, taking a common factor makes calculation shorter.
Step 2
Why this answer is correct
The correct answer is A. (,6,). Taking \(7^4\) common in the numerator gives (\dfrac{74(7-1)}{74}=6). In exams, taking a common factor makes calculation shorter.
Step 3
Exam Tip
ऊपर से \(7^4\) common लेने पर (\dfrac{74(7-1)}{74}=6) मिलता है। परीक्षा में समान factor common लेना गणना को छोटा करता है।
Here \(D=20^2-4\cdot5\cdot20=0\), so the roots are equal and real. In exams, removing a common factor does not change the nature.
Step 2
Why this answer is correct
The correct answer is A. दो बराबर वास्तविक मूल / Two equal real roots. Here \(D=20^2-4\cdot5\cdot20=0\), so the roots are equal and real. In exams, removing a common factor does not change the nature.
Step 3
Exam Tip
यहां \(D=20^2-4\cdot5\cdot20=0\), इसलिए मूल बराबर वास्तविक हैं। परीक्षा में सामान्य गुणनखंड हटाने पर भी प्रकृति नहीं बदलती।
From \(11x^2-77x=0\), (11x(x-7)=0), so (x=0) and (x=7). In exams, dividing by the variable can miss one root.
Step 2
Why this answer is correct
The correct answer is A. (11x(x-7)=0) लिखना / Write (11x(x-7)=0). From \(11x^2-77x=0\), (11x(x-7)=0), so (x=0) and (x=7). In exams, dividing by the variable can miss one root.
Step 3
Exam Tip
\(11x^2-77x=0\) से (11x(x-7)=0), इसलिए (x=0) और (x=7) हैं। परीक्षा में चर से भाग देने से एक मूल छूट सकता है।
From \(9x^2-45x=0\), (9x(x-5)=0), so (x=0) and (x=5). In exams, dividing by the variable can miss one root.
Step 2
Why this answer is correct
The correct answer is A. (9x(x-5)=0) लिखना / Write (9x(x-5)=0). From \(9x^2-45x=0\), (9x(x-5)=0), so (x=0) and (x=5). In exams, dividing by the variable can miss one root.
Step 3
Exam Tip
\(9x^2-45x=0\) से (9x(x-5)=0), इसलिए (x=0) और (x=5) हैं। परीक्षा में चर से भाग देने से एक मूल छूट सकता है।
From \(7x^2-28x=0\), (7x(x-4)=0), so (x=0) and (x=4). In exams, dividing by the variable can miss one root.
Step 2
Why this answer is correct
The correct answer is A. (7x(x-4)=0) लिखना / Write (7x(x-4)=0). From \(7x^2-28x=0\), (7x(x-4)=0), so (x=0) and (x=4). In exams, dividing by the variable can miss one root.
Step 3
Exam Tip
\(7x^2-28x=0\) से (7x(x-4)=0), इसलिए (x=0) और (x=4) हैं। परीक्षा में चर से भाग देने से एक मूल छूट सकता है।
The common factor in \(x^2-16x\) is (x), so we write (x(x-16)=0). In exams, do not divide by the variable and lose (x=0).
Step 2
Why this answer is correct
The correct answer is A. (x(x-16)=0) लिखना / Write (x(x-16)=0). The common factor in \(x^2-16x\) is (x), so we write (x(x-16)=0). In exams, do not divide by the variable and lose (x=0).
Step 3
Exam Tip
\(x^2-16x\) में सामान्य गुणनखंड (x) है, इसलिए (x(x-16)=0) लिखते हैं। परीक्षा में चर से भाग देकर (x=0) को न छोड़ें।
Dividing every term by (8) gives \(x^2-4x+3=0\). Dividing by a common nonzero factor does not change the roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4x+3=0\). Dividing every term by (8) gives \(x^2-4x+3=0\). Dividing by a common nonzero factor does not change the roots.
Step 3
Exam Tip
हर पद को (8) से भाग देने पर \(x^2-4x+3=0\) मिलता है। समान अशून्य गुणनखंड से भाग देने पर मूल नहीं बदलते।
Dividing every term by (6) gives \(x^2-3x+2=0\). Dividing by a common nonzero factor does not change the roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-3x+2=0\). Dividing every term by (6) gives \(x^2-3x+2=0\). Dividing by a common nonzero factor does not change the roots.
Step 3
Exam Tip
हर पद को (6) से भाग देने पर \(x^2-3x+2=0\) मिलता है। समान अशून्य गुणनखंड से भाग देने पर मूल नहीं बदलते।
Dividing every term by (3) gives \(x^2-4x+4=0\). Dividing by a common nonzero factor does not change the roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4x+4=0\). Dividing every term by (3) gives \(x^2-4x+4=0\). Dividing by a common nonzero factor does not change the roots.
Step 3
Exam Tip
हर पद को (3) से भाग देने पर \(x^2-4x+4=0\) मिलता है। समान गुणनखंड से भाग देने पर मूल नहीं बदलते।
After removing the common factor, we get \(x^2-6x+7\), and (D=36-28=8). Since (D) is positive and not a perfect square, the zeroes are real irrational.
Step 2
Why this answer is correct
The correct answer is B. वास्तविक और अपरिमेय / Real and irrational. After removing the common factor, we get \(x^2-6x+7\), and (D=36-28=8). Since (D) is positive and not a perfect square, the zeroes are real irrational.
Step 3
Exam Tip
सामान्य गुणनखंड हटाने पर \(x^2-6x+7\) मिलता है और (D=36-28=8)। (D) धनात्मक अपूर्ण वर्ग है, इसलिए शून्यक वास्तविक अपरिमेय हैं।
Since (3x-2-12x+6=3\(x^2-4x+2\)), the zeroes are \(2\pm\sqrt{2}\). Removing a common factor first makes calculation easier.
Step 2
Why this answer is correct
The correct answer is A. \(2\pm\sqrt{2}\). Since (3x-2-12x+6=3\(x^2-4x+2\)), the zeroes are \(2\pm\sqrt{2}\). Removing a common factor first makes calculation easier.
Step 3
Exam Tip
(3x-2-12x+6=3\(x^2-4x+2\)), इसलिए शून्यक \(2\pm\sqrt{2}\) हैं। पहले सामान्य गुणनखंड हटाना गणना आसान करता है।
A. पहले \(5\mid a\), फिर (a=5k) रखने से \(5\mid b\)/First \(5\mid a\), then substituting (a=5k) gives \(5\mid b\)
Step 1
Concept
From \(a^2=5b^2\), \(5\mid a\).
Step 2
Why this answer is correct
Putting (a=5k) gives \(b^2=5k^2\), so \(5\mid b\).
Step 3
Exam Tip
Now (5) becomes a common factor and gives the contradiction. चरण 1: \(a^2=5b^2\) से \(5\mid a\) मिलता है। चरण 2: (a=5k) रखने पर \(b^2=5k^2\), इसलिए \(5\mid b\)। चरण 3: अब (5) दोनों में साझा गुणनखंड बनकर विरोधाभास देता है।
C. \(2\mid p\) और \(2\mid q\)/\(2\mid p\) and \(2\mid q\)
Step 1
Concept
Coprime numbers have no common factor except (1).
Step 2
Why this answer is correct
\(2\mid p\) and \(2\mid q\) make (2) a common factor.
Step 3
Exam Tip
This is the final contradiction. चरण 1: सहअभाज्य संख्याओं में (1) के अलावा कोई साझा गुणनखंड नहीं होता। चरण 2: \(2\mid p\) और \(2\mid q\) से (2) साझा गुणनखंड बनता है। चरण 3: यही अंतिम विरोधाभास है।
