(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।
Here (ac=600) and (-30+(-20)=-50), so the correct split is (-30x-20x). In exams, even for large (ac), match both sum and product.
Step 2
Why this answer is correct
The correct answer is A. \(24x^2-30x-20x+25=0\). Here (ac=600) and (-30+(-20)=-50), so the correct split is (-30x-20x). In exams, even for large (ac), match both sum and product.
Step 3
Exam Tip
यहां (ac=600) और (-30+(-20)=-50), इसलिए सही विभाजन (-30x-20x) है। परीक्षा में बड़ा (ac) हो तो भी योग और गुणनफल दोनों मिलाएं।
(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।
For equal roots, (D=0), so ((k-4)2=k-2-25) and \(k=\frac{41}{8}\). In exams, handle constant terms carefully while expanding squares.
Step 2
Why this answer is correct
The correct answer is A. \(k=\frac{41}{8}\). For equal roots, (D=0), so ((k-4)2=k-2-25) and \(k=\frac{41}{8}\). In exams, handle constant terms carefully while expanding squares.
Step 3
Exam Tip
समान मूलों के लिए (D=0), इसलिए ((k-4)2=k-2-25) और \(k=\frac{41}{8}\) है। परीक्षा में वर्ग फैलाते समय स्थिर पद ध्यान से लें।
First \(x^2-4x+\frac{7}{11}=0\) is obtained, then ((x-2)2=\frac{37}{11}). In exams, divide by (a) first when \(a\neq1\).
Step 2
Why this answer is correct
The correct answer is A. ((x-2)2=\frac{37}{11}). First \(x^2-4x+\frac{7}{11}=0\) is obtained, then ((x-2)2=\frac{37}{11}). In exams, divide by (a) first when \(a\neq1\).
Step 3
Exam Tip
पहले \(x^2-4x+\frac{7}{11}=0\) बनता है, फिर ((x-2)2=\frac{37}{11}) मिलता है। परीक्षा में \(a\neq1\) हो तो पहले (a) से भाग दें।
Since ((x-2)2=\frac{37}{11}), \(x=2\pm\sqrt{\frac{37}{11}}=2\pm\frac{\sqrt{407}}{11}\). In exams, rationalize the denominator.
Step 2
Why this answer is correct
The correct answer is A. \(x=2\pm\frac{\sqrt{407}}{11}\). Since ((x-2)2=\frac{37}{11}), \(x=2\pm\sqrt{\frac{37}{11}}=2\pm\frac{\sqrt{407}}{11}\). In exams, rationalize the denominator.
Step 3
Exam Tip
((x-2)2=\frac{37}{11}), इसलिए \(x=2\pm\sqrt{\frac{37}{11}}=2\pm\frac{\sqrt{407}}{11}\) है। परीक्षा में हर को परिमेय बनाएं।
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।
(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।
Since (13=\(\sqrt{13}\)2) and the middle term is \(2\sqrt{13}x\), it is (\(x+\sqrt{13}\)2). In exams, identify perfect squares even with irrational coefficients.
Step 2
Why this answer is correct
The correct answer is A. (\(x+\sqrt{13}\)2=0). Since (13=\(\sqrt{13}\)2) and the middle term is \(2\sqrt{13}x\), it is (\(x+\sqrt{13}\)2). In exams, identify perfect squares even with irrational coefficients.
Step 3
Exam Tip
क्योंकि (13=\(\sqrt{13}\)2) और मध्य पद \(2\sqrt{13}x\) है, इसलिए यह (\(x+\sqrt{13}\)2) है। परीक्षा में अपरिमेय गुणांक में भी पूर्ण वर्ग पहचानें।
The equation is equivalent to ((x-s)(x-t)=0), so the roots are (s) 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=s,t). The equation is equivalent to ((x-s)(x-t)=0), so the roots are (s) and (t). In exams, apply zero product rule to symbolic factors too.
Step 3
Exam Tip
यह समीकरण ((x-s)(x-t)=0) के बराबर है, इसलिए मूल (s) और (t) हैं। परीक्षा में प्रतीकात्मक गुणनखंडों पर भी शून्य गुणनफल नियम लगाएं।
Dividing the whole equation by (36) gives (x-2-(m+n)x+mn=0). In exams, removing the common factor first shortens the solution.
Step 2
Why this answer is correct
The correct answer is A. (x=m,n). Dividing the whole equation by (36) gives (x-2-(m+n)x+mn=0). In exams, removing the common factor first shortens the solution.
Step 3
Exam Tip
पूरे समीकरण को (36) से भाग देने पर (x-2-(m+n)x+mn=0) मिलता है। परीक्षा में पहले सामान्य गुणक हटाना हल को छोटा करता है।
Since (x-4=\(x^2\)2=y-2), the new equation is \(y^2-20y+64=0\). In exams, use substitution to form a quadratic.
Step 2
Why this answer is correct
The correct answer is A. \(y^2-20y+64=0\). Since (x-4=\(x^2\)2=y-2), the new equation is \(y^2-20y+64=0\). In exams, use substitution to form a quadratic.
Step 3
Exam Tip
क्योंकि (x-4=\(x^2\)2=y-2), इसलिए नया समीकरण \(y^2-20y+64=0\) है। परीक्षा में प्रतिस्थापन से द्विघात रूप बनाएं।
From \(y^2-20y+64=0\), (y=4,16), so \(x^2=4,16\) and \(x=\pm2,\pm4\). In exams, do not forget to return to (x).
Step 2
Why this answer is correct
The correct answer is B. \(x=\pm4,\pm2\). From \(y^2-20y+64=0\), (y=4,16), so \(x^2=4,16\) and \(x=\pm2,\pm4\). In exams, do not forget to return to (x).
Step 3
Exam Tip
\(y^2-20y+64=0\) से (y=4,16), इसलिए \(x^2=4,16\) और \(x=\pm2,\pm4\) हैं। परीक्षा में वापस (x) के मान निकालना न भूलें।
Multiplying both sides by (6x) gives \(6+6x^2=37x\), that is \(6x^2-37x+6=0\). In exams, remember the condition \(x\neq0\).
Step 2
Why this answer is correct
The correct answer is A. \(6x^2-37x+6=0\). Multiplying both sides by (6x) gives \(6+6x^2=37x\), that is \(6x^2-37x+6=0\). In exams, remember the condition \(x\neq0\).
Step 3
Exam Tip
दोनों पक्षों को (6x) से गुणा करने पर \(6+6x^2=37x\), यानी \(6x^2-37x+6=0\) मिलता है। परीक्षा में \(x\neq0\) शर्त याद रखें।
(6x-2-37x+6=(6x-1)(x-6)), so \(x=\frac{1}{6}\) and (6). In exams, check whether obtained roots are valid in the original equation.
Step 2
Why this answer is correct
The correct answer is A. \(x=6,\frac{1}{6}\). (6x-2-37x+6=(6x-1)(x-6)), so \(x=\frac{1}{6}\) and (6). In exams, check whether obtained roots are valid in the original equation.
Step 3
Exam Tip
(6x-2-37x+6=(6x-1)(x-6)), इसलिए \(x=\frac{1}{6}\) और (6) हैं। परीक्षा में प्राप्त हल मूल समीकरण में मान्य हैं या नहीं जांचें।
Cross multiplication gives ((x+5)2=36x), so \(x^2+10x+25-36x=0\), and \(x^2-26x+25=0\). In exams, cross multiply carefully.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-26x+25=0\). Cross multiplication gives ((x+5)2=36x), so \(x^2+10x+25-36x=0\), and \(x^2-26x+25=0\). In exams, cross multiply carefully.
Step 3
Exam Tip
क्रॉस गुणा करने पर ((x+5)2=36x), इसलिए \(x^2+10x+25-36x=0\) और \(x^2-26x+25=0\) है। परीक्षा में क्रॉस गुणा सावधानी से करें।
(D=(-12)2-4(1)(11)=100), so \(x=\frac{12\pm10}{2}\) gives (1) and (11). In exams, if (D) is a perfect square, simplify quickly.
Step 2
Why this answer is correct
The correct answer is A. (x=1,11). (D=(-12)2-4(1)(11)=100), so \(x=\frac{12\pm10}{2}\) gives (1) and (11). In exams, if (D) is a perfect square, simplify quickly.
Step 3
Exam Tip
(D=(-12)2-4(1)(11)=100), इसलिए \(x=\frac{12\pm10}{2}\) से (1) और (11) मिलते हैं। परीक्षा में (D) पूर्ण वर्ग हो तो उत्तर जल्दी सरल करें।
The sum of roots is \(-\frac{p}{8}\), so \(-\frac{p}{8}=-9\) gives (p=72). In exams, remember the sum formula \(-\frac{b}{a}\).
Step 2
Why this answer is correct
The correct answer is A. (72). The sum of roots is \(-\frac{p}{8}\), so \(-\frac{p}{8}=-9\) gives (p=72). In exams, remember the sum formula \(-\frac{b}{a}\).
Step 3
Exam Tip
मूलों का योग \(-\frac{p}{8}\) है, इसलिए \(-\frac{p}{8}=-9\) से (p=72) है। परीक्षा में योग का सूत्र \(-\frac{b}{a}\) याद रखें।
The product of roots is \(\frac{p}{9}\), so \(\frac{p}{9}=\frac{1}{3}\) 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 \(\frac{p}{9}\), so \(\frac{p}{9}=\frac{1}{3}\) gives (p=3). In exams, use the product formula \(\frac{c}{a}\).
Step 3
Exam Tip
मूलों का गुणनफल \(\frac{p}{9}\) है, इसलिए \(\frac{p}{9}=\frac{1}{3}\) से (p=3) है। परीक्षा में गुणनफल का सूत्र \(\frac{c}{a}\) लगाएं।
\(\alpha+\beta=23\) and \(\alpha\beta=126\), so (\alpha-2+\beta-2=232-2(126)=277). In exams, remember (\(\alpha+\beta\)2-2\alpha\beta).
Step 2
Why this answer is correct
The correct answer is A. (277). \(\alpha+\beta=23\) and \(\alpha\beta=126\), so (\alpha-2+\beta-2=232-2(126)=277). In exams, remember (\(\alpha+\beta\)2-2\alpha\beta).
Step 3
Exam Tip
\(\alpha+\beta=23\) और \(\alpha\beta=126\), इसलिए (\alpha-2+\beta-2=232-2(126)=277) है। परीक्षा में (\(\alpha+\beta\)2-2\alpha\beta) याद रखें।
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{20}{91}\). In exams, first write sum and product in reciprocal questions.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{20}{91}\). \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{20}{91}\). In exams, first write sum and product in reciprocal questions.
Step 3
Exam Tip
\(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha\beta}=\frac{20}{91}\) होता है। परीक्षा में reciprocal वाले प्रश्न में पहले योग और गुणनफल लिखें।
The sum of roots is \(-\frac{b}{a}=-\frac{-25}{7}=\frac{25}{7}\). In exams, keep the sign of (b) carefully.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{25}{7}\). The sum of roots is \(-\frac{b}{a}=-\frac{-25}{7}=\frac{25}{7}\). In exams, keep the sign of (b) carefully.
Step 3
Exam Tip
मूलों का योग \(-\frac{b}{a}=-\frac{-25}{7}=\frac{25}{7}\) है। परीक्षा में (b) का चिन्ह ध्यान से रखें।
(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{25}{7}\right\)2-\frac{48}{7}=\frac{289}{49}). In exams, convert fractions to a common denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{289}{49}\). (\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{25}{7}\right\)2-\frac{48}{7}=\frac{289}{49}). In exams, convert fractions to a common denominator.
Step 3
Exam Tip
(\(\alpha-\beta\)2=\(\alpha+\beta\)2-4\alpha\beta=\left\(\frac{25}{7}\right\)2-\frac{48}{7}=\frac{289}{49}) है। परीक्षा में भिन्नों को समान हर में बदलें।
(\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=10\) and \(\alpha\beta=-24\), so the value is (-240). In exams, factor the expression first.
Step 2
Why this answer is correct
The correct answer is A. (-240). (\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), where \(\alpha+\beta=10\) and \(\alpha\beta=-24\), so the value is (-240). In exams, factor the expression first.
Step 3
Exam Tip
(\alpha-2\beta+\alpha\beta-2=\alpha\beta\(\alpha+\beta\)), जहां \(\alpha+\beta=10\) और \(\alpha\beta=-24\), इसलिए मान (-240) है। परीक्षा में अभिव्यक्ति को पहले factor करें।
Let the roots be (-r) and (-2r), then \(2r^2=49\) and \(p=3r=\frac{21\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{21\sqrt{2}}{2}\). Let the roots be (-r) and (-2r), then \(2r^2=49\) and \(p=3r=\frac{21\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 3
Exam Tip
मूलों को (-r) और (-2r) मानें, तो \(2r^2=49\) और \(p=3r=\frac{21\sqrt{2}}{2}\) है। परीक्षा में हर को परिमेय बनाना न भूलें।
Here (D=(-11)2-4(1)(6)=97), so the difference of roots is \(\frac{\sqrt{D}}{|a|}=\sqrt{97}\). In exams, the difference of roots can be found directly from (D).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{97}\). Here (D=(-11)2-4(1)(6)=97), so the difference of roots is \(\frac{\sqrt{D}}{|a|}=\sqrt{97}\). In exams, the difference of roots can be found directly from (D).
Step 3
Exam Tip
यहां (D=(-11)2-4(1)(6)=97), इसलिए मूलों का अंतर \(\frac{\sqrt{D}}{|a|}=\sqrt{97}\) है। परीक्षा में मूलों का अंतर सीधे (D) से मिल सकता है।
Here (D=(-12)2-4(6)(17)=-264<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=(-12)2-4(6)(17)=-264<0), so there are no real roots. In exams, (D<0) means no real roots.
Step 3
Exam Tip
यहां (D=(-12)2-4(6)(17)=-264<0), इसलिए वास्तविक मूल नहीं हैं। परीक्षा में (D<0) का अर्थ वास्तविक मूल नहीं है।
(6x-2-12x+17=6(x-1)2+11), 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. (6(x-1)2+11=0). (6x-2-12x+17=6(x-1)2+11), so it cannot be zero for real (x). In exams, completed square form also shows the nature of roots.
Step 3
Exam Tip
(6x-2-12x+17=6(x-1)2+11), 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 (3,11), so new roots are (8,16), and the equation is ((x-8)(x-16)=0). In exams, form the new roots and then the new equation.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-24x+128=0\). The roots are (3,11), so new roots are (8,16), and the equation is ((x-8)(x-16)=0). In exams, form the new roots and then the new equation.
Step 3
Exam Tip
मूल (3,11) हैं, इसलिए नए मूल (8,16) होंगे और समीकरण ((x-8)(x-16)=0) है। परीक्षा में नए मूल बनाकर नया समीकरण लिखें।
\(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=22\) and \(\alpha\beta=117\), so the value is \(\frac{484-234}{117}=\frac{250}{117}\). In exams, convert expressions into sum and product.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{250}{117}\). \(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}=\frac{\alpha^2+\beta^2}{\alpha\beta}\), where \(\alpha+\beta=22\) and \(\alpha\beta=117\), so the value is \(\frac{484-234}{117}=\frac{250}{117}\). 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 \(\alpha+\beta=22\) और \(\alpha\beta=117\), इसलिए मान \(\frac{484-234}{117}=\frac{250}{117}\) है। परीक्षा में अभिव्यक्ति को योग और गुणनफल में बदलें।
((x-6)(x-13)=x-2-19x+78), so \(x^2-19x+78=22\) gives \(x^2-19x+56=0\). In exams, bring all terms to one side after expansion.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-19x+56=0\). ((x-6)(x-13)=x-2-19x+78), so \(x^2-19x+78=22\) gives \(x^2-19x+56=0\). In exams, bring all terms to one side after expansion.
Step 3
Exam Tip
((x-6)(x-13)=x-2-19x+78), इसलिए \(x^2-19x+78=22\) से \(x^2-19x+56=0\) मिलता है। परीक्षा में विस्तार के बाद सभी पद एक तरफ लाएं।
Here (D=(-19)2-4(1)(56)=137), so \(x=\frac{19\pm\sqrt{137}}{2}\). In exams, finding (D) correctly is important.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{19\pm\sqrt{137}}{2}\). Here (D=(-19)2-4(1)(56)=137), so \(x=\frac{19\pm\sqrt{137}}{2}\). In exams, finding (D) correctly is important.
Step 3
Exam Tip
यहां (D=(-19)2-4(1)(56)=137), इसलिए \(x=\frac{19\pm\sqrt{137}}{2}\) है। परीक्षा में (D) को सही निकालना जरूरी है।
First \(x^2+6x+\frac{5}{3}=0\) is obtained, then adding (9) gives ((x+3)2=\frac{22}{3}). In exams, divide by (a) first when \(a\neq1\).
Step 2
Why this answer is correct
The correct answer is A. ((x+3)2=\frac{22}{3}). First \(x^2+6x+\frac{5}{3}=0\) is obtained, then adding (9) gives ((x+3)2=\frac{22}{3}). In exams, divide by (a) first when \(a\neq1\).
Step 3
Exam Tip
पहले \(x^2+6x+\frac{5}{3}=0\) मिलता है, फिर (9) जोड़ने पर ((x+3)2=\frac{22}{3}) बनता है। परीक्षा में \(a\neq1\) हो तो पहले (a) से भाग दें।
