(7x-2-14x+19=7(x-1)2+12), 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. (7(x-1)2+12=0). (7x-2-14x+19=7(x-1)2+12), so it cannot be zero for real (x). In exams, completed square form also shows the nature of roots.
Step 3
Exam Tip
(7x-2-14x+19=7(x-1)2+12), इसलिए यह वास्तविक (x) के लिए शून्य नहीं हो सकता। परीक्षा में पूर्ण वर्ग रूप से भी मूलों की प्रकृति दिखती है।
Since ((x-2)2=\frac{43}{13}), \(x=2\pm\sqrt{\frac{43}{13}}=2\pm\frac{\sqrt{559}}{13}\). In exams, rationalize the denominator.
Step 2
Why this answer is correct
The correct answer is A. \(x=2\pm\frac{\sqrt{559}}{13}\). Since ((x-2)2=\frac{43}{13}), \(x=2\pm\sqrt{\frac{43}{13}}=2\pm\frac{\sqrt{559}}{13}\). In exams, rationalize the denominator.
Step 3
Exam Tip
((x-2)2=\frac{43}{13}), इसलिए \(x=2\pm\sqrt{\frac{43}{13}}=2\pm\frac{\sqrt{559}}{13}\) है। परीक्षा में हर को परिमेय बनाएं।
First \(x^2-4x+\frac{9}{13}=0\) is obtained, then ((x-2)2=\frac{43}{13}). 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{43}{13}). First \(x^2-4x+\frac{9}{13}=0\) is obtained, then ((x-2)2=\frac{43}{13}). In exams, divide by (a) first when \(a\neq1\).
Step 3
Exam Tip
पहले \(x^2-4x+\frac{9}{13}=0\) बनता है, फिर ((x-2)2=\frac{43}{13}) मिलता है। परीक्षा में \(a\neq1\) हो तो पहले (a) से भाग दें।
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।
(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।
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।
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।
(5x-2-10x+13=5(x-1)2+8), 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. (5(x-1)2+8=0). (5x-2-10x+13=5(x-1)2+8), so it cannot be zero for real (x). In exams, completed square form also shows the nature of roots.
Step 3
Exam Tip
(5x-2-10x+13=5(x-1)2+8), इसलिए यह वास्तविक (x) के लिए शून्य नहीं हो सकता। परीक्षा में पूर्ण वर्ग रूप से भी मूलों की प्रकृति दिखती है।
Since (\left\(x-\frac{5}{3}\right\)2=\frac{17}{9}), \(x=\frac{5\pm\sqrt{17}}{3}\). In exams, write the square root with the denominator correctly.
Step 2
Why this answer is correct
The correct answer is A. \(x=\frac{5\pm\sqrt{17}}{3}\). Since (\left\(x-\frac{5}{3}\right\)2=\frac{17}{9}), \(x=\frac{5\pm\sqrt{17}}{3}\). In exams, write the square root with the denominator correctly.
Step 3
Exam Tip
(\left\(x-\frac{5}{3}\right\)2=\frac{17}{9}), इसलिए \(x=\frac{5\pm\sqrt{17}}{3}\) है। परीक्षा में वर्गमूल को हर के साथ सही लिखें।
First \(x^2-\frac{10}{3}x+\frac{8}{9}=0\) is obtained, then (\left\(x-\frac{5}{3}\right\)2=\frac{17}{9}). In exams, divide by (a) first when \(a\neq1\).
Step 2
Why this answer is correct
The correct answer is A. (\left\(x-\frac{5}{3}\right\)2=\frac{17}{9}). First \(x^2-\frac{10}{3}x+\frac{8}{9}=0\) is obtained, then (\left\(x-\frac{5}{3}\right\)2=\frac{17}{9}). In exams, divide by (a) first when \(a\neq1\).
Step 3
Exam Tip
पहले \(x^2-\frac{10}{3}x+\frac{8}{9}=0\) बनता है, फिर (\left\(x-\frac{5}{3}\right\)2=\frac{17}{9}) मिलता है। परीक्षा में \(a\neq1\) हो तो पहले (a) से भाग दें।
(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।
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।
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।
First \(x^2+5x+\frac{3}{2}=0\) is obtained, then adding \(\frac{25}{4}\) gives (\left\(x+\frac{5}{2}\right\)2=\frac{19}{4}). In exams, divide by (a) first when \(a\neq1\).
Step 2
Why this answer is correct
The correct answer is A. (\left\(x+\frac{5}{2}\right\)2=\frac{19}{4}). First \(x^2+5x+\frac{3}{2}=0\) is obtained, then adding \(\frac{25}{4}\) gives (\left\(x+\frac{5}{2}\right\)2=\frac{19}{4}). In exams, divide by (a) first when \(a\neq1\).
Step 3
Exam Tip
पहले \(x^2+5x+\frac{3}{2}=0\) मिलता है, फिर \(\frac{25}{4}\) जोड़ने पर (\left\(x+\frac{5}{2}\right\)2=\frac{19}{4}) बनता है। परीक्षा में \(a\neq1\) हो तो पहले (a) से भाग दें।
(3x-2-6x+7=3(x-1)2+4), so it cannot be zero for real (x). In exams, completed square form also shows the nature of real roots.
Step 2
Why this answer is correct
The correct answer is A. (3(x-1)2+4=0). (3x-2-6x+7=3(x-1)2+4), so it cannot be zero for real (x). In exams, completed square form also shows the nature of real roots.
Step 3
Exam Tip
(3x-2-6x+7=3(x-1)2+4), इसलिए यह वास्तविक (x) के लिए शून्य नहीं हो सकता। परीक्षा में पूर्ण वर्ग रूप से भी वास्तविक मूलों की प्रकृति दिखती है।
First \(x^2-\frac{18}{5}x+\frac{9}{5}=0\) is obtained, then (\left\(x-\frac{9}{5}\right\)2=\frac{36}{25}). In exams, divide by (a) first when \(a\neq1\).
Step 2
Why this answer is correct
The correct answer is A. (\left\(x-\frac{9}{5}\right\)2=\frac{36}{25}). First \(x^2-\frac{18}{5}x+\frac{9}{5}=0\) is obtained, then (\left\(x-\frac{9}{5}\right\)2=\frac{36}{25}). In exams, divide by (a) first when \(a\neq1\).
Step 3
Exam Tip
पहले \(x^2-\frac{18}{5}x+\frac{9}{5}=0\) बनता है, फिर (\left\(x-\frac{9}{5}\right\)2=\frac{36}{25}) मिलता है। परीक्षा में \(a\neq1\) हो तो पहले (a) से भाग दें।
Since ((x+2)2=\frac{7}{2}), \(x=-2\pm\sqrt{\frac{7}{2}}=-2\pm\frac{\sqrt{14}}{2}\). In exams, write the square root in simplified form.
Step 2
Why this answer is correct
The correct answer is A. \(x=-2\pm\frac{\sqrt{14}}{2}\). Since ((x+2)2=\frac{7}{2}), \(x=-2\pm\sqrt{\frac{7}{2}}=-2\pm\frac{\sqrt{14}}{2}\). In exams, write the square root in simplified form.
Step 3
Exam Tip
((x+2)2=\frac{7}{2}), इसलिए \(x=-2\pm\sqrt{\frac{7}{2}}=-2\pm\frac{\sqrt{14}}{2}\) है। परीक्षा में वर्गमूल को सरल रूप में लिखें।
First \(x^2+4x+\frac{1}{2}=0\) is obtained, then adding (4) gives ((x+2)2=\frac{7}{2}). 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{7}{2}). First \(x^2+4x+\frac{1}{2}=0\) is obtained, then adding (4) gives ((x+2)2=\frac{7}{2}). In exams, divide by (a) first when \(a\neq1\).
Step 3
Exam Tip
पहले \(x^2+4x+\frac{1}{2}=0\) मिलता है, फिर (4) जोड़ने पर ((x+2)2=\frac{7}{2}) बनता है। परीक्षा में \(a\neq1\) हो तो पहले (a) से भाग दें।
(2x-2-4x+5=2(x-1)2+3), so it cannot be zero for real (x). In exams, completed square form also shows no real roots.
Step 2
Why this answer is correct
The correct answer is A. (2(x-1)2+3=0). (2x-2-4x+5=2(x-1)2+3), so it cannot be zero for real (x). In exams, completed square form also shows no real roots.
Step 3
Exam Tip
(2x-2-4x+5=2(x-1)2+3), इसलिए यह वास्तविक (x) के लिए शून्य नहीं हो सकता। परीक्षा में पूर्ण वर्ग रूप से भी no real roots दिखता है।
