Last updated at June 29, 2018 by Teachoo
Transcript
Example 14 Find the roots of the following equations: (i) x + 1/๐ฅ=3,๐ฅโ 0 x + 1/๐ฅ=3 (๐ฅ (๐ฅ) + 1)/๐ฅ=3 (๐ฅ2 + 1)/๐ฅ=3 x2 + 1 = 3x x2 โ 3x + 1 = 0 We will factorize by quadratic formula Comparing equation with ax2 + bx + c = 0 Here, a = 1, b = โ3, c = 1 We know that D = b2 โ 4ac D = (โ3)2 โ 4 (1) (1) D = 9 โ 4 D = 5 So, the roots of the equation is given by x = (โ ๐ ยฑ โ๐ท)/2๐ Putting values x = (โ (โ3) ยฑ โ5)/(2 ร 1) x = (3 ยฑ โ5)/2 Hence, the roots of the equation are (3 + โ5)/2 & (3 โ โ5)/2 , Example 14 Find the roots of the following equations: (ii) 1/๐ฅโ1/(๐ฅโ2)=3,๐ฅโ 0,2 1/๐ฅโ1/(๐ฅ โ 2)=3 ((๐ฅ โ 2) โ ๐ฅ )/(๐ฅ(๐ฅ โ 2))=3 (โ2 )/(๐ฅ(๐ฅ โ 2))=3 โ2 = 3x(x โ 2) โ2 = 3x2 โ 6x 0 = 3x2 โ 6x + 2 3x2 โ 6x + 2 = 0 We solve this equation by quadratic formula 3x2 โ 6x + 2 = 0 Comparing equation with ax2 + bx + c = 0 Here, a = 3, b = โ6, c = 2 We know that D = b2 โ 4ac D = (โ 6)2 โ 4ร(3)ร(2) D = 36 โ 24 D = 12 So, the roots of the equation is given by x = (โ ๐ ยฑ โ๐ท)/2๐ Putting values x = (โ(โ 6) ยฑ โ12)/(2 ร 3) x = (6 ยฑ โ12)/6 x = (6 ยฑ โ(4 ร 3))/6 x = (6 ยฑ โ(4 ) รโ3)/6 x = (6 ยฑ 2 โ3)/6 x = (2(3 ยฑ โ3))/(2 ร 3) x = (3 ยฑ โ3)/3 So , the roots of the equation are (3 + โ3)/3 and (3 โ โ3)/3
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Example 1 (ii)
Example 2 Important
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Example 7 Deleted for CBSE Board 2022 Exams
Example 8 Important Deleted for CBSE Board 2022 Exams
Example 9 Deleted for CBSE Board 2022 Exams
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Example 14 You are here
Example 15 Important
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Example 18 Important
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