# Ex 4.3, 1

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 4.3 ,1 Find the roots of the following quadratic equations, if they exist, by the method of completing the square: (i) 2x2 – 7x + 3 = 0 2x2 – 7x +3 = 0 Dividing by 2 (2𝑥2 − 7𝑥 + 3 = 0)/2=0/2 2𝑥2/2 – 7𝑥/2 + 3/2=0 x2 – 7𝑥/2+3/2=0 We know that (a – b)2 = a2 – 2ab + b2 Here, a = x & – 2ab = – 7𝑥/2 – 2xb = −7𝑥/2 b = −7𝑥/(2(−2𝑥)) b = 7/4 Now, in our equation x2 −7𝑥/2+3/2=0 Adding and subtracting (7/4)^2 x2 −7𝑥/2+3/2+(7/4)^2−(7/4)^2=0 x2 +(7/4)^2−7𝑥/2+3/2−(7/4)^2=0 (𝑥− 7/4)^2+3/2 −(7/4)^2=0 (𝑥− 7/4)^2+3/2−49/16=0 (𝑥− 7/4)^2+(3(8) − 49)/16=0 (𝑥− 7/4)^2+(24 − 49)/16=0 (𝑥− 7/4)^2−25/16=0 (𝑥−7/4)^2=25/16 (𝑥−7/4)^2=(5/4)^2 Cancelling square both sides 𝑥−7/4 = ± 5/4 Ex 4.3 ,1 Find the roots of the following quadratic equations, if they exist, by the method of completing the square: (ii) 2x2 + x – 4 = 0 2x2 + x – 4 = 0 Dividing whole equation by 2 (2𝑥2 + 𝑥 − 4)/2=0/2 2𝑥2/2+𝑥/2−4/2=0 x2 + 𝑥/2−2=0 We know that (a + b)2 = a2 + 2ab + b2 Here, a = x & 2ab = 𝑥/2 2xb = 𝑥/2 2b = 1/2 b = 1/2 × 1/2 b = 1/4 Now, in our equation x2 + 𝑥/2−2=0 Adding and subtracting (1/4)^2 x2 + 𝑥/2−2+(1/4)^2−(1/4)^2= 0 x2 + 𝑥/2+(1/4)^2– 2 – (1/4)^2=0 (𝑥+1/4)^2−2−(1/4)^2= 0 (𝑥+1/4)^2−2 −1/16=0 (𝑥+1/4)^2=2+1/16 (𝑥+1/4)^2=(2(16) + 1)/16 (𝑥+1/4)^2=(32 + 1)/16 (𝑥+1/4)^2=33/16 (𝑥+1/4)^2=(√33/4)^2 Cancelling square both sides 𝑥+1/4 = ± √33/4 Solving So, the root of the equation are x = (√33 − 1)/4 & x = (−(√33 + 1))/4 Ex 4.3 ,1 Find the roots of the following quadratic equations, if they exist, by the method of completing the square: (iii) 4x2 + 4√3 𝑥+3=0 4x2 + 4 √3 𝑥+3=0 Dividing whole equation 4 (4𝑥^2+ 4 √3 𝑥+ 3)/4=0/4 (4𝑥^2)/4 + (4 √3)/4 x + 3/4=0 x2 + √3 𝑥+ 3/4 = 0 We know that (a + b)2 = a2 + 2ab + b2 Here, a = x & 2ab = √3 𝑥 2xb = √3 𝑥 2b = √3 b = √3/2 Now, in our equation x2 + √3 𝑥+3/4=0 Adding and subtracting (√3/2)^2 x2 + √3 𝑥+3/4+(√3/2)^2−(√3/2)^2=0 x2 + √3 𝑥+(√3/2)^2+3/4−(√3/2)^2=0 (𝑥+√3/2 )^2+3/4−(√3/2)^2=0 (𝑥+√3/2 )^2+3/4−3/4=0 (𝑥+√3/2 )^2=0 (𝑥+√3/2 )^2=02 Cancelling square both sides (𝑥+√3/2 )^2= ± 0 So, the root of the equation are x = (−√3)/2 & x = (−√3)/2 Ex 4.3 ,1 Find the roots of the following quadratic equations, if they exist, by the method of completing the square: (iv) 2x2 + x + 4 = 0 2x2 + x + 4 = 0 Dividing equation by 2 (2𝑥2 + 𝑥 + 4)/2=0/2 2𝑥2/2+𝑥/2 + 4/2=0 x2 + 𝑥/2+2=0 We know that (a + b)2 = a2 + 2ab + b2 Here, a = x & 2ab = 𝑥/2 2xb = 𝑥/2 2b = 1/2 b = 1/2×1/2 b = 1/4 Now, in our equation x2 + 𝑥/2+2=0 Adding and subtracting (1/4)^2 x2 + 𝑥/2 +2+(1/4)^2−(1/4)^2=0 "x2 + " 𝑥/2+(1/4)^2+2−(1/4)^2=0 (𝑥+1/4)^2+2−(1/4)^2=0 (𝑥+1/4)^2+2−1/16=0 (𝑥+1/4)^2+((32 − 1)/16)=0 (𝑥+1/4)^2+ 31/16=0 (𝑥+1/4)^2=(−31)/16 Since square of any number cannot be negative So, answer does not exist

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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 .