(x-2+14x+48=(x+6)(x+8)), so (x=-6,-8). In exams, a positive middle term and positive constant can give negative roots.
Step 2
Why this answer is correct
The correct answer is A. (x=-6,-8). (x-2+14x+48=(x+6)(x+8)), so (x=-6,-8). In exams, a positive middle term and positive constant can give negative roots.
Step 3
Exam Tip
(x-2+14x+48=(x+6)(x+8)), इसलिए (x=-6,-8) हैं। परीक्षा में धनात्मक मध्य पद और धनात्मक स्थिर पद से ऋणात्मक मूल आ सकते हैं।
(8x-2-2x-3=(4x-3)(2x+1)), so the roots are \(\frac{3}{4}\) and \(-\frac{1}{2}\). In exams, solve both linear factors carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{4},-\frac{1}{2}\). (8x-2-2x-3=(4x-3)(2x+1)), so the roots are \(\frac{3}{4}\) and \(-\frac{1}{2}\). In exams, solve both linear factors carefully.
Step 3
Exam Tip
(8x-2-2x-3=(4x-3)(2x+1)), इसलिए मूल \(\frac{3}{4}\) और \(-\frac{1}{2}\) हैं। परीक्षा में दोनों रैखिक गुणनखंड सावधानी से हल करें।
(5x-2+16x+3=(5x+1)(x+3)), so the roots are \(-\frac{1}{5}\) and (-3). In exams, positive factors give negative roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=-3,-\frac{1}{5}\). (5x-2+16x+3=(5x+1)(x+3)), so the roots are \(-\frac{1}{5}\) and (-3). In exams, positive factors give negative roots.
Step 3
Exam Tip
(5x-2+16x+3=(5x+1)(x+3)), इसलिए मूल \(-\frac{1}{5}\) और (-3) हैं। परीक्षा में धनात्मक गुणनखंडों से ऋणात्मक मूल मिलते हैं।
The roots of the first equation are (5,6), and the roots of the second are (6,7). In exams, solve both equations separately and compare the common root.
Step 2
Why this answer is correct
The correct answer is A. (x=6). The roots of the first equation are (5,6), and the roots of the second are (6,7). In exams, solve both equations separately and compare the common root.
Step 3
Exam Tip
पहले समीकरण के मूल (5,6) और दूसरे के मूल (6,7) हैं। परीक्षा में दोनों समीकरण अलग हल करके समान मूल देखें।
(5x-2+9x-2=(5x-1)(x+2)), so the roots are \(\frac{1}{5}\) and (-2). In exams, correct standard form is very important.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{1}{5},-2\). (5x-2+9x-2=(5x-1)(x+2)), so the roots are \(\frac{1}{5}\) and (-2). In exams, correct standard form is very important.
Step 3
Exam Tip
(5x-2+9x-2=(5x-1)(x+2)), इसलिए मूल \(\frac{1}{5}\) और (-2) हैं। परीक्षा में सही मानक रूप बहुत जरूरी है।
(4x-2-19x-5=(4x+1)(x-5)), so the roots are (5) and \(-\frac{1}{4}\). In exams, reverse the signs from linear factors to write roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=5,-\frac{1}{4}\). (4x-2-19x-5=(4x+1)(x-5)), so the roots are (5) and \(-\frac{1}{4}\). In exams, reverse the signs from linear factors to write roots.
Step 3
Exam Tip
(4x-2-19x-5=(4x+1)(x-5)), इसलिए मूल (5) और \(-\frac{1}{4}\) हैं। परीक्षा में रैखिक गुणनखंडों के चिन्ह उलटकर मूल लिखें।
(12x-2+17x+5=(12x+5)(x+1)), so the roots are \(-\frac{5}{12}\) and (-1). In exams, positive factors give negative roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=-1,-\frac{5}{12}\). (12x-2+17x+5=(12x+5)(x+1)), so the roots are \(-\frac{5}{12}\) and (-1). In exams, positive factors give negative roots.
Step 3
Exam Tip
(12x-2+17x+5=(12x+5)(x+1)), इसलिए मूल \(-\frac{5}{12}\) और (-1) हैं। परीक्षा में धनात्मक गुणनखंडों से ऋणात्मक मूल मिलते हैं।
(7x-2-9x+2=(7x-2)(x-1)), so the roots are (1) and \(\frac{2}{7}\). In exams, solve each linear factor separately.
Step 2
Why this answer is correct
The correct answer is A. \(x=1,\frac{2}{7}\). (7x-2-9x+2=(7x-2)(x-1)), so the roots are (1) and \(\frac{2}{7}\). In exams, solve each linear factor separately.
Step 3
Exam Tip
(7x-2-9x+2=(7x-2)(x-1)), इसलिए मूल (1) और \(\frac{2}{7}\) हैं। परीक्षा में हर रैखिक गुणनखंड को अलग हल करें।
(x-2+12x+32=(x+4)(x+8)), so (x=-4,-8). In exams, a positive middle term and positive constant can give negative roots.
Step 2
Why this answer is correct
The correct answer is A. (x=-4,-8). (x-2+12x+32=(x+4)(x+8)), so (x=-4,-8). In exams, a positive middle term and positive constant can give negative roots.
Step 3
Exam Tip
(x-2+12x+32=(x+4)(x+8)), इसलिए (x=-4,-8) हैं। परीक्षा में धनात्मक मध्य पद और धनात्मक स्थिर पद से ऋणात्मक मूल आ सकते हैं।
(6x-2+x-2=(3x+2)(2x-1)), so the roots are \(\frac{1}{2}\) and \(-\frac{2}{3}\). In exams, solve both linear factors carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{1}{2},-\frac{2}{3}\). (6x-2+x-2=(3x+2)(2x-1)), so the roots are \(\frac{1}{2}\) and \(-\frac{2}{3}\). In exams, solve both linear factors carefully.
Step 3
Exam Tip
(6x-2+x-2=(3x+2)(2x-1)), इसलिए मूल \(\frac{1}{2}\) और \(-\frac{2}{3}\) हैं। परीक्षा में दोनों रैखिक गुणनखंड सावधानी से हल करें।
(3x-2+11x+10=(3x+5)(x+2)), so the roots are \(-\frac{5}{3}\) and (-2). In exams, positive factors give negative roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=-2,-\frac{5}{3}\). (3x-2+11x+10=(3x+5)(x+2)), so the roots are \(-\frac{5}{3}\) and (-2). In exams, positive factors give negative roots.
Step 3
Exam Tip
(3x-2+11x+10=(3x+5)(x+2)), इसलिए मूल \(-\frac{5}{3}\) और (-2) हैं। परीक्षा में धनात्मक गुणनखंडों से ऋणात्मक मूल मिलते हैं।
The roots of the first equation are (3,4), and the roots of the second are (4,5). In exams, solve both equations separately and compare the common root.
Step 2
Why this answer is correct
The correct answer is A. (x=4). The roots of the first equation are (3,4), and the roots of the second are (4,5). In exams, solve both equations separately and compare the common root.
Step 3
Exam Tip
पहले समीकरण के मूल (3,4) और दूसरे के मूल (4,5) हैं। परीक्षा में दोनों समीकरण अलग हल करके समान मूल देखें।
(3x-2+7x-6=(3x-2)(x+3)), so the roots are \(\frac{2}{3}\) and (-3). In exams, correct standard form is necessary first.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{2}{3},-3\). (3x-2+7x-6=(3x-2)(x+3)), so the roots are \(\frac{2}{3}\) and (-3). In exams, correct standard form is necessary first.
Step 3
Exam Tip
(3x-2+7x-6=(3x-2)(x+3)), इसलिए मूल \(\frac{2}{3}\) और (-3) हैं। परीक्षा में पहले सही मानक रूप बनाना जरूरी है।
(5x-2-13x-6=(5x+2)(x-3)), so the roots are (3) and \(-\frac{2}{5}\). In exams, reverse the signs from linear factors to write roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=3,-\frac{2}{5}\). (5x-2-13x-6=(5x+2)(x-3)), so the roots are (3) and \(-\frac{2}{5}\). In exams, reverse the signs from linear factors to write roots.
Step 3
Exam Tip
(5x-2-13x-6=(5x+2)(x-3)), इसलिए मूल (3) और \(-\frac{2}{5}\) हैं। परीक्षा में रैखिक गुणनखंडों के चिन्ह उलटकर मूल लिखें।
(3x-2+8x+4=(3x+2)(x+2)), so the roots are \(-\frac{2}{3}\) and (-2). In exams, positive factors give negative roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=-2,-\frac{2}{3}\). (3x-2+8x+4=(3x+2)(x+2)), so the roots are \(-\frac{2}{3}\) and (-2). In exams, positive factors give negative roots.
Step 3
Exam Tip
(3x-2+8x+4=(3x+2)(x+2)), इसलिए मूल \(-\frac{2}{3}\) और (-2) हैं। परीक्षा में धनात्मक गुणनखंडों से ऋणात्मक मूल मिलते हैं।
(8x-2+14x+3=(4x+1)(2x+3)), so the roots are \(-\frac{1}{4}\) and \(-\frac{3}{2}\). In exams, write fractional roots in simplest form.
Step 2
Why this answer is correct
The correct answer is A. \(x=-\frac{1}{4},-\frac{3}{2}\). (8x-2+14x+3=(4x+1)(2x+3)), so the roots are \(-\frac{1}{4}\) and \(-\frac{3}{2}\). In exams, write fractional roots in simplest form.
Step 3
Exam Tip
(8x-2+14x+3=(4x+1)(2x+3)), इसलिए मूल \(-\frac{1}{4}\) और \(-\frac{3}{2}\) हैं। परीक्षा में भिन्न मूल सरल रूप में लिखें।
(4x-2-12x+5=(2x-1)(2x-5)), so the roots are \(\frac{1}{2}\) and \(\frac{5}{2}\). In exams, solve each linear factor separately.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{1}{2},\frac{5}{2}\). (4x-2-12x+5=(2x-1)(2x-5)), so the roots are \(\frac{1}{2}\) and \(\frac{5}{2}\). In exams, solve each linear factor separately.
Step 3
Exam Tip
(4x-2-12x+5=(2x-1)(2x-5)), इसलिए मूल \(\frac{1}{2}\) और \(\frac{5}{2}\) हैं। परीक्षा में हर रैखिक गुणनखंड को अलग हल करें।
(x-2+10x+21=(x+3)(x+7)), so (x=-3,-7). In exams, a positive middle term and positive constant can give both negative roots.
Step 2
Why this answer is correct
The correct answer is A. (x=-3,-7). (x-2+10x+21=(x+3)(x+7)), so (x=-3,-7). In exams, a positive middle term and positive constant can give both negative roots.
Step 3
Exam Tip
(x-2+10x+21=(x+3)(x+7)), इसलिए (x=-3,-7) हैं। परीक्षा में धनात्मक मध्य पद और धनात्मक स्थिर पद पर दोनों मूल ऋणात्मक हो सकते हैं।
(2x-2+7x+6=(2x+3)(x+2)), so \(x=-\frac{3}{2}\) and (-2). In exams, positive factors give negative roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=-\frac{3}{2},-2\). (2x-2+7x+6=(2x+3)(x+2)), so \(x=-\frac{3}{2}\) and (-2). In exams, positive factors give negative roots.
Step 3
Exam Tip
(2x-2+7x+6=(2x+3)(x+2)), इसलिए \(x=-\frac{3}{2}\) और (-2) हैं। परीक्षा में धनात्मक गुणनखंडों से ऋणात्मक मूल मिलते हैं।
The roots of the first equation are (2,3), and the roots of the second are (2,4). In exams, solve both equations separately and compare roots.
Step 2
Why this answer is correct
The correct answer is A. (x=2). The roots of the first equation are (2,3), and the roots of the second are (2,4). In exams, solve both equations separately and compare roots.
Step 3
Exam Tip
पहले समीकरण के मूल (2,3) और दूसरे के मूल (2,4) हैं। परीक्षा में दोनों समीकरण अलग-अलग हल करके समान मूल देखें।
