# Example 13

Last updated at Dec. 8, 2016 by Teachoo

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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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .