Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 13 Find the roots of the following quadratic equations, if they exist, using the quadratic formula: (i) 3x2 – 5x + 2 = 0 3x2 – 5x + 2 = 0 Comparing equation with ax2 + bx + c = 0 Here, a = 3, b = – 5, c = 2 We know that, D = b2 – 4ac D = (–5)2 – 4×(3)×(2) D = 25 – 24 D = 1 So, the roots of the equation is given by x = (− 𝑏 ± √𝐷)/2𝑎 Putting values x = (−(− 5)± √1)/(2 × 3) x = (5 ± 1)/6 Solving Hence, the roots of the equation are 1 and 2/3 . Example 13 Find the roots of the following quadratic equations, if they exist, using the quadratic formula: (ii) x2 + 4x + 5 = 0 x2 + 4x + 5 = 0 Comparing equation with ax2 + bx + c = 0 Here, a = 1, b = 4, c = 5 We know that D = b2 – 4ac D = (4)2 – 4×(1) (5) D = (4)2 – 20 D = 16 – 20 D = –4 Hence roots to equation are x = (− 𝑏 ± √𝐷)/2𝑎 x = (− 4 ± √(− 4))/(2 (1)) Since there is a negative number in the root, therefore √D will not have any real value. So, there are no real roots for the given equation. Example 13 Find the roots of the following quadratic equations, if they exist, using the quadratic formula: (iii) 2x2 – 2 √2 𝑥 + 1 = 0 2x2 - 2√2 x + 1 = 0 Comparing equation with ax2 + bx + c = 0 Here, a = 2, b = – 2√2 , c = 1 We know that D = b2 – 4ac = ("– 2" √2)^2– 4×(2)×1 = (4×2)−(8) = 8 – 8 = 0 So, the roots of the equation is given by x = (− 𝑏 ± √𝐷)/2𝑎 Putting values x = (− (− 2√2) ± √0)/(2 × 2) x = (2√2 ± 0)/4 x = (2√2)/4 x = √2/2 Hence, the roots of the equation are √2/2 and √2/2.

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.