(10x-2+x-3=(5x+3)(2x-1)), so \(x=\frac{1}{2},-\frac{3}{5}\) is correct. In exams, change signs carefully from factors.
Step 2
Why this answer is correct
The correct answer is A. सही है / Correct. (10x-2+x-3=(5x+3)(2x-1)), so \(x=\frac{1}{2},-\frac{3}{5}\) is correct. In exams, change signs carefully from factors.
Step 3
Exam Tip
(10x-2+x-3=(5x+3)(2x-1)), इसलिए \(x=\frac{1}{2},-\frac{3}{5}\) सही है। परीक्षा में गुणनखंडों से संकेत सावधानी से बदलें।
The first equation has roots \(\frac{3}{2},\frac{9}{4}\), and the second has roots \(\frac{3}{2},\frac{10}{9}\). In exams, solve both equations separately for the common root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{2}\). The first equation has roots \(\frac{3}{2},\frac{9}{4}\), and the second has roots \(\frac{3}{2},\frac{10}{9}\). In exams, solve both equations separately for the common root.
Step 3
Exam Tip
पहले समीकरण के मूल \(\frac{3}{2},\frac{9}{4}\) और दूसरे के मूल \(\frac{3}{2},\frac{10}{9}\) हैं। परीक्षा में समान मूल के लिए दोनों समीकरण अलग हल करें।
(24x-2-50x+25=(6x-5)(4x-5)), so the roots are \(\frac{5}{6}\) and \(\frac{5}{4}\). In exams, keep the denominator coefficients correctly.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{5}{6},\frac{5}{4}\). (24x-2-50x+25=(6x-5)(4x-5)), so the roots are \(\frac{5}{6}\) and \(\frac{5}{4}\). In exams, keep the denominator coefficients correctly.
Step 3
Exam Tip
(24x-2-50x+25=(6x-5)(4x-5)), इसलिए मूल \(\frac{5}{6}\) और \(\frac{5}{4}\) हैं। परीक्षा में हर वाले गुणांक को सही रखें।
(16x-2-38x+15=(8x-3)(2x-5)), so the roots are \(\frac{3}{8}\) and \(\frac{5}{2}\). In exams, do not invert fractional roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{8},\frac{5}{2}\). (16x-2-38x+15=(8x-3)(2x-5)), so the roots are \(\frac{3}{8}\) and \(\frac{5}{2}\). In exams, do not invert fractional roots.
Step 3
Exam Tip
(16x-2-38x+15=(8x-3)(2x-5)), इसलिए मूल \(\frac{3}{8}\) और \(\frac{5}{2}\) हैं। परीक्षा में भिन्न मूलों को उल्टा न लिखें।
((3x+2)(2x-5)=0), so \(x=-\frac{2}{3}\) and \(\frac{5}{2}\). In exams, change signs while writing roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{5}{2},-\frac{2}{3}\). ((3x+2)(2x-5)=0), so \(x=-\frac{2}{3}\) and \(\frac{5}{2}\). In exams, change signs while writing roots.
Step 3
Exam Tip
((3x+2)(2x-5)=0), इसलिए \(x=-\frac{2}{3}\) और \(\frac{5}{2}\) हैं। परीक्षा में संकेत बदलकर मूल लिखें।
The equation is equivalent to ((x-r)(x-t)=0), so the roots are (r) and (t). In exams, apply zero product rule to symbolic factors too.
Step 2
Why this answer is correct
The correct answer is A. (x=r,t). The equation is equivalent to ((x-r)(x-t)=0), so the roots are (r) and (t). In exams, apply zero product rule to symbolic factors too.
Step 3
Exam Tip
यह समीकरण ((x-r)(x-t)=0) के बराबर है, इसलिए मूल (r) और (t) हैं। परीक्षा में प्रतीकात्मक गुणनखंडों पर भी शून्य गुणनफल नियम लगाएं।
(6x-2+x-2=(3x+2)(2x-1)), so \(x=\frac{1}{2},-\frac{2}{3}\) is correct. In exams, change signs carefully from factors.
Step 2
Why this answer is correct
The correct answer is A. सही है / Correct. (6x-2+x-2=(3x+2)(2x-1)), so \(x=\frac{1}{2},-\frac{2}{3}\) is correct. In exams, change signs carefully from factors.
Step 3
Exam Tip
(6x-2+x-2=(3x+2)(2x-1)), इसलिए \(x=\frac{1}{2},-\frac{2}{3}\) सही है। परीक्षा में गुणनखंडों से संकेत सावधानी से बदलें।
The first equation has roots \(\frac{3}{2},\frac{5}{3}\), and the second has roots \(\frac{3}{2},\frac{6}{5}\). In exams, solve both equations separately for the common root.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{2}\). The first equation has roots \(\frac{3}{2},\frac{5}{3}\), and the second has roots \(\frac{3}{2},\frac{6}{5}\). In exams, solve both equations separately for the common root.
Step 3
Exam Tip
पहले समीकरण के मूल \(\frac{3}{2},\frac{5}{3}\) और दूसरे के मूल \(\frac{3}{2},\frac{6}{5}\) हैं। परीक्षा में समान मूल के लिए दोनों समीकरण अलग हल करें।
(20x-2-43x+21=(5x-7)(4x-3)), so the roots are \(\frac{7}{5}\) and \(\frac{3}{4}\). In exams, do not invert fractional roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{7}{5},\frac{3}{4}\). (20x-2-43x+21=(5x-7)(4x-3)), so the roots are \(\frac{7}{5}\) and \(\frac{3}{4}\). In exams, do not invert fractional roots.
Step 3
Exam Tip
(20x-2-43x+21=(5x-7)(4x-3)), इसलिए मूल \(\frac{7}{5}\) और \(\frac{3}{4}\) हैं। परीक्षा में भिन्न मूलों को उल्टा न लिखें।
(14x-2-25x+6=(7x-3)(2x-2)), so the roots are \(\frac{3}{7}\) and (1). In exams, also check by removing any common factor if present.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{7},1\). (14x-2-25x+6=(7x-3)(2x-2)), so the roots are \(\frac{3}{7}\) and (1). In exams, also check by removing any common factor if present.
Step 3
Exam Tip
(14x-2-25x+6=(7x-3)(2x-2)), इसलिए मूल \(\frac{3}{7}\) और (1) हैं। परीक्षा में पहले सामान्य गुणक हो तो उसे हटाकर भी जांचें।
The equation is equivalent to ((x-u)(x-v)=0), so the roots are (u) and (v). In exams, apply the same rule to symbolic factors.
Step 2
Why this answer is correct
The correct answer is A. (x=u,v). The equation is equivalent to ((x-u)(x-v)=0), so the roots are (u) and (v). In exams, apply the same rule to symbolic factors.
Step 3
Exam Tip
यह समीकरण ((x-u)(x-v)=0) के बराबर है, इसलिए मूल (u) और (v) हैं। परीक्षा में प्रतीकात्मक गुणनखंडों पर भी वही नियम लागू करें।
(4x-2+4x-3=(2x-1)(2x+3)), so \(x=\frac{1}{2},-\frac{3}{2}\) is correct. In exams, change signs carefully from factors.
Step 2
Why this answer is correct
The correct answer is A. सही है / Correct. (4x-2+4x-3=(2x-1)(2x+3)), so \(x=\frac{1}{2},-\frac{3}{2}\) is correct. In exams, change signs carefully from factors.
Step 3
Exam Tip
(4x-2+4x-3=(2x-1)(2x+3)), इसलिए \(x=\frac{1}{2},-\frac{3}{2}\) सही है। परीक्षा में गुणनखंडों से संकेत सावधानी से बदलें।
(18x-2-27x+10=(3x-2)(6x-5)), so the roots are \(\frac{2}{3}\) and \(\frac{5}{6}\). In exams, write fractional roots in simplest form.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{2}{3},\frac{5}{6}\). (18x-2-27x+10=(3x-2)(6x-5)), so the roots are \(\frac{2}{3}\) and \(\frac{5}{6}\). In exams, write fractional roots in simplest form.
Step 3
Exam Tip
(18x-2-27x+10=(3x-2)(6x-5)), इसलिए मूल \(\frac{2}{3}\) और \(\frac{5}{6}\) हैं। परीक्षा में भिन्न मूलों को सरल रूप में लिखें।
(10x-2-13x+3=(10x-3)(x-1)), so the roots are (1) and \(\frac{3}{10}\). In exams, set each linear factor equal to zero.
Step 2
Why this answer is correct
The correct answer is A. \(x=1,\frac{3}{10}\). (10x-2-13x+3=(10x-3)(x-1)), so the roots are (1) and \(\frac{3}{10}\). In exams, set each linear factor equal to zero.
Step 3
Exam Tip
(10x-2-13x+3=(10x-3)(x-1)), इसलिए मूल (1) और \(\frac{3}{10}\) हैं। परीक्षा में हर रैखिक गुणनखंड को अलग शून्य रखें।
The equation is equivalent to ((x-m)(x-n)=0), so the roots are (m) and (n). In exams, the same rule applies to symbolic factors.
Step 2
Why this answer is correct
The correct answer is A. (x=m,n). The equation is equivalent to ((x-m)(x-n)=0), so the roots are (m) and (n). In exams, the same rule applies to symbolic factors.
Step 3
Exam Tip
यह समीकरण ((x-m)(x-n)=0) के बराबर है, इसलिए मूल (m) और (n) हैं। परीक्षा में प्रतीकात्मक गुणनखंडों पर भी वही नियम लागू होता है।
(15x-2+16x+4=(3x+2)(5x+2)), so the roots are \(-\frac{2}{3}\) and \(-\frac{2}{5}\). In exams, write fractional roots in simplest form.
Step 2
Why this answer is correct
The correct answer is A. \(x=-\frac{2}{3},-\frac{2}{5}\). (15x-2+16x+4=(3x+2)(5x+2)), so the roots are \(-\frac{2}{3}\) and \(-\frac{2}{5}\). In exams, write fractional roots in simplest form.
Step 3
Exam Tip
(15x-2+16x+4=(3x+2)(5x+2)), इसलिए मूल \(-\frac{2}{3}\) और \(-\frac{2}{5}\) हैं। परीक्षा में भिन्न मूलों को सरल रूप में लिखें।
(8x-2-14x+3=(4x-1)(2x-3)), so the roots are \(\frac{1}{4}\) and \(\frac{3}{2}\). In exams, set each linear factor equal to zero.
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, set each linear factor equal to zero.
Step 3
Exam Tip
(8x-2-14x+3=(4x-1)(2x-3)), इसलिए मूल \(\frac{1}{4}\) और \(\frac{3}{2}\) हैं। परीक्षा में रैखिक गुणनखंडों को अलग-अलग शून्य रखें।
(12x-2-17x+6=(3x-2)(4x-3)), so the roots are \(\frac{2}{3}\) and \(\frac{3}{4}\). In exams, write fractional roots in simplest form.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{2}{3},\frac{3}{4}\). (12x-2-17x+6=(3x-2)(4x-3)), so the roots are \(\frac{2}{3}\) and \(\frac{3}{4}\). In exams, write fractional roots in simplest form.
Step 3
Exam Tip
(12x-2-17x+6=(3x-2)(4x-3)), इसलिए मूल \(\frac{2}{3}\) और \(\frac{3}{4}\) हैं। परीक्षा में भिन्न मूलों को सरल रूप में लिखें।
(6x-2-7x-3=(3x+1)(2x-3)), so the roots are \(-\frac{1}{3}\) and \(\frac{3}{2}\). In exams, solve both linear factors separately.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{2},-\frac{1}{3}\). (6x-2-7x-3=(3x+1)(2x-3)), so the roots are \(-\frac{1}{3}\) and \(\frac{3}{2}\). In exams, solve both linear factors separately.
Step 3
Exam Tip
(6x-2-7x-3=(3x+1)(2x-3)), इसलिए मूल \(-\frac{1}{3}\) और \(\frac{3}{2}\) हैं। परीक्षा में दोनों रैखिक गुणनखंड अलग-अलग हल करें।
((2x+1)(2x-7)=0), so \(x=-\frac{1}{2}\) and \(\frac{7}{2}\). In exams, change signs while writing roots.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{7}{2},-\frac{1}{2}\). ((2x+1)(2x-7)=0), so \(x=-\frac{1}{2}\) and \(\frac{7}{2}\). In exams, change signs while writing roots.
Step 3
Exam Tip
((2x+1)(2x-7)=0), इसलिए \(x=-\frac{1}{2}\) और \(\frac{7}{2}\) हैं। परीक्षा में संकेत बदलकर मूल लिखें।
