Last updated at Dec. 8, 2016 by Teachoo

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Ex 4.3 ,2 Find the roots of the quadratic equation using quadratic formula (i) 2x2 – 7x + 3 = 0 2x2 – 7x + 3 = 0 Comparing equation with ax2 + bx + c = 0 a = 2, b = – 7, c = 3 We know that D = b2 – 4ac D = ( – 7)2 – 4×2×3 D = ( –7 ×−7)−(4×2×3) D = 49 – 24 D = 25 The roots to equation is given by x = (− 𝑏 ± √𝐷)/2𝑎 Putting values x = (− (− 7) ± √25)/(2 × 2) x = (7 ± √(5^2 ))/4 x = (7 ± 5)/4 Solving Both Hence , the roots to equation are 3 and 1/2 . Ex 4.3 ,2 Find the roots of the quadratic equation using quadratic formula (ii) 2x2 + x – 4 = 0 2x2 + x – 4 = 0 Comparing equation with ax2 + bx + c = 0 a = 2, b = 1, c = – 4 We know that D = b2 – 4ac D = 12 – 4 ×2×−4 D = 1 + 32 = 33 So, the roots to equation are x = (−𝑏 ± √𝐷)/2𝑎 Putting values x = (−1 ± √33)/(2 × 2) x = (−1 ± √33)/4 Hence x = (−1 + √33)/4 & x = (−1 − √33)/4 are the roots of the equation Ex 4.3 ,2 Find the roots of the quadratic equation using quadratic formula (iii) 4x2 + 4√3 𝑥+3=0 4x2 + 4√3 𝑥+3=0 Comparing equation with ax2 + bx + c = 0 a = 4, b = 4√3, c = 3 We know that, D = b2 – 4ac D = (4 √3)2 – 4×4×3 D = 4√3×4√3−4×4×3 D = 16 ×3−4×4×3 D = 48 – 48 D = 0 Hence , roots to equation are x = (−𝑏 ± √𝐷)/2𝑎 Putting values x = (− (4√3) ± √0)/(2 × 4) x = (−(4√3) )/(2 × 4) x = (−√3)/2 Therefore x = (−√3)/2 & x = (−√3)/2 are the roots of the equation Ex 4.3 ,2 Find the roots of the quadratic equation using quadratic formula (iv) 2x2 + x + 4 = 0 2x2 + x + 4 = 0 Comparing equation with ax2 + bx + c = 0 So, a = 2, b = 1, c = 4 We know that D = b2 – 4ac D = 12 – 4×2×4 D = 1 – 32 D = –31 Hence roots to equation are x = (− 𝑏 ± √𝐷)/2𝑎 Putting values x = (− 1 ± √(−31))/2𝑎 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.

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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.