(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।
Here (ac=180) and (-15+(-12)=-27), so the correct split is (-15x-12x). In exams, match both sum (b) and product (ac).
Step 2
Why this answer is correct
The correct answer is A. \(18x^2-15x-12x+10=0\). Here (ac=180) and (-15+(-12)=-27), so the correct split is (-15x-12x). In exams, match both sum (b) and product (ac).
Step 3
Exam Tip
यहां (ac=180) और (-15+(-12)=-27), इसलिए सही विभाजन (-15x-12x) है। परीक्षा में योग (b) और गुणनफल (ac) दोनों मिलाएं।
(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।
For equal roots, (D=0), so (4(k-2)2-4\(k^2-9\)=0) gives (-4k+13=0). In exams, expand (D) carefully in parameter questions.
Step 2
Why this answer is correct
The correct answer is A. \(k=\frac{13}{4}\). For equal roots, (D=0), so (4(k-2)2-4\(k^2-9\)=0) gives (-4k+13=0). In exams, expand (D) carefully in parameter questions.
Step 3
Exam Tip
समान मूलों के लिए (D=0), इसलिए (4(k-2)2-4\(k^2-9\)=0) से (-4k+13=0) मिलता है। परीक्षा में पैरामीटर वाले प्रश्नों में (D) सावधानी से फैलाएं।
First \(x^2-\frac{22}{7}x+1=0\) is obtained, then (\left\(x-\frac{11}{7}\right\)2=\frac{72}{49}). In exams, divide by (a) first when \(a\neq1\).
Step 2
Why this answer is correct
The correct answer is A. (\left\(x-\frac{11}{7}\right\)2=\frac{72}{49}). First \(x^2-\frac{22}{7}x+1=0\) is obtained, then (\left\(x-\frac{11}{7}\right\)2=\frac{72}{49}). In exams, divide by (a) first when \(a\neq1\).
Step 3
Exam Tip
पहले \(x^2-\frac{22}{7}x+1=0\) बनता है, फिर (\left\(x-\frac{11}{7}\right\)2=\frac{72}{49}) मिलता है। परीक्षा में \(a\neq1\) हो तो पहले (a) से भाग दें।
Since (\left\(x-\frac{11}{7}\right\)2=\frac{72}{49}), \(x=\frac{11\pm6\sqrt{2}}{7}\). In exams, simplify \(\sqrt{72}=6\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{11\pm6\sqrt{2}}{7}\). Since (\left\(x-\frac{11}{7}\right\)2=\frac{72}{49}), \(x=\frac{11\pm6\sqrt{2}}{7}\). In exams, simplify \(\sqrt{72}=6\sqrt{2}\).
Step 3
Exam Tip
(\left\(x-\frac{11}{7}\right\)2=\frac{72}{49}), इसलिए \(x=\frac{11\pm6\sqrt{2}}{7}\) है। परीक्षा में \(\sqrt{72}=6\sqrt{2}\) सरल करें।
The first equation has roots \(2,\frac{6}{5}\), and the second has roots \(\frac{3}{2},\frac{4}{3}\), so there is no common root among the given values. In exams, solve both equations before comparing.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{2}\). The first equation has roots \(2,\frac{6}{5}\), and the second has roots \(\frac{3}{2},\frac{4}{3}\), so there is no common root among the given values. In exams, solve both equations before comparing.
Step 3
Exam Tip
पहले समीकरण के मूल \(\frac{6}{5},2\) नहीं बल्कि \(\frac{6}{5}\) और (2) हैं, इसलिए यह विकल्प नहीं है। सही जांच में दोनों समीकरणों के मूल क्रमशः \(2,\frac{6}{5}\) और \(\frac{3}{2},\frac{4}{3}\) हैं।
The first equation has roots \(\frac{1}{2},\frac{5}{2}\), and the second has roots \(\frac{3}{2},\frac{4}{3}\), so none of the listed values is common. In exams, solve both equations correctly before comparing.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{3}{2}\). The first equation has roots \(\frac{1}{2},\frac{5}{2}\), and the second has roots \(\frac{3}{2},\frac{4}{3}\), so none of the listed values is common. In exams, solve both equations correctly before comparing.
Step 3
Exam Tip
पहले समीकरण के मूल \(\frac{1}{2},\frac{5}{2}\) हैं और दूसरे के मूल \(\frac{3}{2},\frac{4}{3}\) हैं, इसलिए दिए विकल्पों में समान मूल नहीं है। परीक्षा में तुलना से पहले दोनों समीकरण सही हल 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।
Since (7=\(\sqrt{7}\)2) and the middle term is \(2\sqrt{7}x\), it is (\(x+\sqrt{7}\)2). In exams, identify perfect squares even with irrational coefficients.
Step 2
Why this answer is correct
The correct answer is A. (\(x+\sqrt{7}\)2=0). Since (7=\(\sqrt{7}\)2) and the middle term is \(2\sqrt{7}x\), it is (\(x+\sqrt{7}\)2). In exams, identify perfect squares even with irrational coefficients.
Step 3
Exam Tip
क्योंकि (7=\(\sqrt{7}\)2) और मध्य पद \(2\sqrt{7}x\) है, इसलिए यह (\(x+\sqrt{7}\)2) है। परीक्षा में अपरिमेय गुणांक में भी पूर्ण वर्ग पहचानें।
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।
Dividing the whole equation by (16) gives (x-2-(a+b)x+ab=0). In exams, removing the common factor first makes solving easier.
Step 2
Why this answer is correct
The correct answer is A. (x=a,b). Dividing the whole equation by (16) gives (x-2-(a+b)x+ab=0). In exams, removing the common factor first makes solving easier.
Step 3
Exam Tip
पूरे समीकरण को (16) से भाग देने पर (x-2-(a+b)x+ab=0) मिलता है। परीक्षा में पहले सामान्य गुणक हटाना हल को आसान करता है।
Since (x-4=\(x^2\)2=y-2), the new equation is \(y^2-13y+36=0\). In exams, substitution simplifies a difficult form.
Step 2
Why this answer is correct
The correct answer is A. \(y^2-13y+36=0\). Since (x-4=\(x^2\)2=y-2), the new equation is \(y^2-13y+36=0\). In exams, substitution simplifies a difficult form.
Step 3
Exam Tip
क्योंकि (x-4=\(x^2\)2=y-2), इसलिए नया समीकरण \(y^2-13y+36=0\) है। परीक्षा में प्रतिस्थापन कठिन रूप को सरल बनाता है।
From \(y^2-13y+36=0\), (y=4,9), so \(x^2=4,9\) and \(x=\pm2,\pm3\). In exams, do not forget to return to (x).
Step 2
Why this answer is correct
The correct answer is A. \(x=\pm2,\pm3\). From \(y^2-13y+36=0\), (y=4,9), so \(x^2=4,9\) and \(x=\pm2,\pm3\). In exams, do not forget to return to (x).
Step 3
Exam Tip
\(y^2-13y+36=0\) से (y=4,9), इसलिए \(x^2=4,9\) और \(x=\pm2,\pm3\) हैं। परीक्षा में वापस (x) के मान निकालना न भूलें।
Multiplying both sides by (4x) gives \(4+4x^2=17x\), that is \(4x^2-17x+4=0\). In exams, remember the condition \(x\neq0\).
Step 2
Why this answer is correct
The correct answer is A. \(4x^2-17x+4=0\). Multiplying both sides by (4x) gives \(4+4x^2=17x\), that is \(4x^2-17x+4=0\). In exams, remember the condition \(x\neq0\).
Step 3
Exam Tip
दोनों पक्षों को (4x) से गुणा करने पर \(4+4x^2=17x\), यानी \(4x^2-17x+4=0\) मिलता है। परीक्षा में \(x\neq0\) शर्त याद रखें।
(4x-2-17x+4=(4x-1)(x-4)), so \(x=\frac{1}{4}\) and (4). In exams, check whether obtained roots are valid in the original equation.
Step 2
Why this answer is correct
The correct answer is A. \(x=4,\frac{1}{4}\). (4x-2-17x+4=(4x-1)(x-4)), so \(x=\frac{1}{4}\) and (4). In exams, check whether obtained roots are valid in the original equation.
Step 3
Exam Tip
(4x-2-17x+4=(4x-1)(x-4)), इसलिए \(x=\frac{1}{4}\) और (4) हैं। परीक्षा में मिले हल मूल समीकरण में मान्य हैं या नहीं जांचें।
Cross multiplication gives ((x+3)2=16x), so \(x^2+6x+9-16x=0\), and \(x^2-10x+9=0\). In exams, cross multiply carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-10x+9=0\). Cross multiplication gives ((x+3)2=16x), so \(x^2+6x+9-16x=0\), and \(x^2-10x+9=0\). In exams, cross multiply carefully.
Step 3
Exam Tip
क्रॉस गुणा करने पर ((x+3)2=16x), इसलिए \(x^2+6x+9-16x=0\) और \(x^2-10x+9=0\) है। परीक्षा में क्रॉस गुणा सावधानी से करें।
(D=(-8)2-4(1)(3)=52), so \(x=\frac{8\pm2\sqrt{13}}{2}=4\pm\sqrt{13}\). In exams, simplify the square root.
Step 2
Why this answer is correct
The correct answer is A. \(x=4\pm\sqrt{13}\). (D=(-8)2-4(1)(3)=52), so \(x=\frac{8\pm2\sqrt{13}}{2}=4\pm\sqrt{13}\). In exams, simplify the square root.
Step 3
Exam Tip
(D=(-8)2-4(1)(3)=52), इसलिए \(x=\frac{8\pm2\sqrt{13}}{2}=4\pm\sqrt{13}\) है। परीक्षा में वर्गमूल को सरल करें।
The sum of roots is \(-\frac{p}{5}\), so \(-\frac{p}{5}=-7\) gives (p=35). In exams, remember the sum formula \(-\frac{b}{a}\).
Step 2
Why this answer is correct
The correct answer is A. (35). The sum of roots is \(-\frac{p}{5}\), so \(-\frac{p}{5}=-7\) gives (p=35). In exams, remember the sum formula \(-\frac{b}{a}\).
Step 3
Exam Tip
मूलों का योग \(-\frac{p}{5}\) है, इसलिए \(-\frac{p}{5}=-7\) से (p=35) है। परीक्षा में योग का सूत्र \(-\frac{b}{a}\) याद रखें।
The product of roots is \(\frac{p}{6}\), so \(\frac{p}{6}=\frac{1}{2}\) gives (p=3). In exams, use the product formula \(\frac{c}{a}\).
Step 2
Why this answer is correct
The correct answer is A. (3). The product of roots is \(\frac{p}{6}\), so \(\frac{p}{6}=\frac{1}{2}\) gives (p=3). In exams, use the product formula \(\frac{c}{a}\).
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{p}{6}\) है, इसलिए \(\frac{p}{6}=\frac{1}{2}\) से (p=3) है। परीक्षा में गुणनफल का सूत्र \(\frac{c}{a}\) लगाएं।
\(\alpha+\beta=17\) and \(\alpha\beta=70\), so (\alpha-2+\beta-2=172-2(70)=149). In exams, remember (\(\alpha+\beta\)2-2\alpha\beta).
Step 2
Why this answer is correct
The correct answer is A. (149). \(\alpha+\beta=17\) and \(\alpha\beta=70\), so (\alpha-2+\beta-2=172-2(70)=149). In exams, remember (\(\alpha+\beta\)2-2\alpha\beta).
Step 3
Exam Tip
\(\alpha+\beta=17\) और \(\alpha\beta=70\), इसलिए (\alpha-2+\beta-2=172-2(70)=149) है। परीक्षा में (\(\alpha+\beta\)2-2\alpha\beta) याद रखें।
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{12}{35}\). In exams, first write sum and product in reciprocal questions.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{12}{35}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{12}{35}\). In exams, first write sum and product in reciprocal questions.
Step 3
Exam Tip
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{12}{35}\) होता है। परीक्षा में reciprocal वाले प्रश्न में पहले योग और गुणनफल लिखें।
The sum of roots is \(-\frac{b}{a}=-\frac{-13}{4}=\frac{13}{4}\). In exams, keep the sign of (b) carefully.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{13}{4}\). The sum of roots is \(-\frac{b}{a}=-\frac{-13}{4}=\frac{13}{4}\). In exams, keep the sign of (b) carefully.
Step 3
Exam Tip
मूलों का योग \(-\frac{b}{a}=-\frac{-13}{4}=\frac{13}{4}\) है। परीक्षा में (b) का चिन्ह ध्यान से रखें।
(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{13}{4}\right\)2-3=\frac{121}{16}). In exams, use this identity for the square of difference.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{121}{16}\). (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{13}{4}\right\)2-3=\frac{121}{16}). In exams, use this identity for the square of difference.
Step 3
Exam Tip
(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{13}{4}\right\)2-3=\frac{121}{16}) है। परीक्षा में अंतर का वर्ग इस पहचान से निकालें।
(\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=6\) and \(\alpha\beta=-16\), so the value is (-96). In exams, factor the expression first.
Step 2
Why this answer is correct
The correct answer is A. (-96). (\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=6\) and \(\alpha\beta=-16\), so the value is (-96). In exams, factor the expression first.
Step 3
Exam Tip
(\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), जहां \(\alpha+\beta=6\) और \(\alpha\beta=-16\), इसलिए मान (-96) है। परीक्षा में अभिव्यक्ति को पहले factor करें।
Let the roots be (-r) and (-2r), then \(2r^2=25\) and \(p=3r=\frac{15\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{15\sqrt{2}}{2}\). Let the roots be (-r) and (-2r), then \(2r^2=25\) and \(p=3r=\frac{15\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=25\) और \(p=3r=\frac{15\sqrt{2}}{2}\) है। परीक्षा में हर को परिमेय बनाना न भूलें।
Here (D=(-7)2-4(1)(3)=37), so the difference of roots is \(\frac{\sqrt{D}}{|a|}=\sqrt{37}\). In exams, the difference of roots can be found directly from (D).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{37}\). Here (D=(-7)2-4(1)(3)=37), so the difference of roots is \(\frac{\sqrt{D}}{|a|}=\sqrt{37}\). In exams, the difference of roots can be found directly from (D).
Step 3
Exam Tip
यहां (D=(-7)2-4(1)(3)=37), इसलिए मूलों का अंतर \(\frac{\sqrt{D}}{|a|}=\sqrt{37}\) है। परीक्षा में मूलों का अंतर सीधे (D) से मिल सकता है।
Here (D=(-8)2-4(4)(9)=-80<0), so there are no real roots. In exams, (D<0) means no real roots.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक मूल नहीं हैं / There are no real roots. Here (D=(-8)2-4(4)(9)=-80<0), so there are no real roots. In exams, (D<0) means no real roots.
Step 3
Exam Tip
यहां (D=(-8)2-4(4)(9)=-80<0), इसलिए वास्तविक मूल नहीं हैं। परीक्षा में (D<0) का अर्थ वास्तविक मूल नहीं है।
(4x-2-8x+9=4(x-1)2+5), so it cannot be zero for real (x). In exams, completed square form also shows the nature of roots.
Step 2
Why this answer is correct
The correct answer is A. (4(x-1)2+5=0). (4x-2-8x+9=4(x-1)2+5), so it cannot be zero for real (x). In exams, completed square form also shows the nature of roots.
Step 3
Exam Tip
(4x-2-8x+9=4(x-1)2+5), इसलिए यह वास्तविक (x) के लिए शून्य नहीं हो सकता। परीक्षा में पूर्ण वर्ग रूप से भी मूलों की प्रकृति दिखती है।
The roots are (2,8), so new roots are (5,11), and the equation is ((x-5)(x-11)=0). In exams, form the new roots and then the new equation.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-16x+55=0\). The roots are (2,8), so new roots are (5,11), and the equation is ((x-5)(x-11)=0). In exams, form the new roots and then the new equation.
Step 3
Exam Tip
मूल (2,8) हैं, इसलिए नए मूल (5,11) होंगे और समीकरण ((x-5)(x-11)=0) है। परीक्षा में नए मूल बनाकर नया समीकरण लिखें।
\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=14\) and \(\alpha\beta=45\), so the value is \(\frac{196-90}{45}=\frac{106}{45}\). In exams, convert expressions into sum and product.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{106}{45}\). \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=14\) and \(\alpha\beta=45\), so the value is \(\frac{196-90}{45}=\frac{106}{45}\). In exams, convert expressions into sum and product.
Step 3
Exam Tip
\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), जहां \(\alpha+\beta=14\) और \(\alpha\beta=45\), इसलिए मान \(\frac{196-90}{45}=\frac{106}{45}\) है। परीक्षा में अभिव्यक्ति को योग और गुणनफल में बदलें।
((x-4)(x-9)=x-2-13x+36), so \(x^2-13x+36=14\) gives \(x^2-13x+22=0\). In exams, bring all terms to one side after expansion.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-13x+22=0\). ((x-4)(x-9)=x-2-13x+36), so \(x^2-13x+36=14\) gives \(x^2-13x+22=0\). In exams, bring all terms to one side after expansion.
Step 3
Exam Tip
((x-4)(x-9)=x-2-13x+36), इसलिए \(x^2-13x+36=14\) से \(x^2-13x+22=0\) मिलता है। परीक्षा में विस्तार के बाद सभी पद एक तरफ लाएं।
Here (D=(-13)2-4(1)(22)=81), so \(x=\frac{13\pm9}{2}\). In exams, if (D) is a perfect square, the answer simplifies quickly.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{13\pm9}{2}\). Here (D=(-13)2-4(1)(22)=81), so \(x=\frac{13\pm9}{2}\). In exams, if (D) is a perfect square, the answer simplifies quickly.
Step 3
Exam Tip
यहां (D=(-13)2-4(1)(22)=81), इसलिए \(x=\frac{13\pm9}{2}\) है। परीक्षा में (D) पूर्ण वर्ग हो तो उत्तर जल्दी सरल होता है।
