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Transcript

Ex 4.3, 1 Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: (iii) 2x2 – 6x + 3 = 0 2x2 – 6x + 3 = 0 Comparing equation with ax2 + bx + c = 0 a = 2, b = –6 , c = 3 We know that D = b2 – 4ac = (–6)2 – 4 × 𝟐 × 𝟑 = (–6) × (–6) – 4 × 2 × 3 = 36 – 24 = 12 Since D > 0 There are 2 distinct real roots Now using quadratic formula to find roots x = (− 𝑏 ± √𝐷)/2𝑎 Putting values x = (−(− 𝟔) ± √𝟏𝟐)/(𝟐 × 𝟐) x = (+ 6 ± √12)/4 x = (6 ± √(4 × 3))/4 x = (6 ± √(4 )× √3)/4 x = (𝟔 ± 𝟐 ×√𝟑)/𝟒 x = (2(3 ± √3))/4 x = (𝟑 ± √𝟑)/𝟐 Therefore, the root of the equation are x = (3 + √3)/2 & x = (3 − √3)/2

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.