Last updated at May 29, 2018 by Teachoo

Transcript

Ex 4.4 ,1 Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: (i) 2x2 – 3x + 5 = 0 2x2 – 3x + 5 = 0 Comparing equation with ax2 + bx + c = 0 a = 2 , b = –3, c = 5 We know that , D = b2 – 4ac = (–3)2 – 4 × 2 × 5 = 9 – 40 = – 31 Since D < 0 Hence, there are no real roots Ex 4.4 ,1 Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: (ii) 3x2 – 4 √3 x + 4 = 0 3x2 – 4√3 x + 4 = 0 Comparing equation with ax2 + bx + c = 0 a = 3, b = – 4√3, c = 4 We know that D = b2 – 4ac D = ( – 4√3 )2 – 4 ×3×4 D = (− 4√3×−4√3)−4×3×4 D = (− 4×− 4×√3×√3)−4×3×4 D = 16 ×3−4×3×4 D = 48 – 48 D = 0 Since D = 0 The given equation has two equal real roots Now using quadratic formula to find roots x = (− 𝑏 ± √𝐷)/2𝑎 Putting the values x = (−(− 4√3) ± √0)/(2 × 3) x = 4(√3+0)/6 x = 4(√3)/6 x = (2√3)/3 Hence, x = (2√3)/3 & x = (2√3)/3 are the roots of the equation Ex 4.4 ,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

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.