Last updated at May 29, 2018 by Teachoo

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Ex 4.3 ,1 Find the roots of the following quadratic equations, if they exist, by the method of completing the square: (i) 2x2 โ 7x + 3 = 0 2x2 โ 7x +3 = 0 Dividing by 2 (2๐ฅ2 โ 7๐ฅ + 3 = 0)/2=0/2 2๐ฅ2/2 โ 7๐ฅ/2 + 3/2=0 x2 โ 7๐ฅ/2+3/2=0 We know that (a โ b)2 = a2 โ 2ab + b2 Here, a = x & โ 2ab = โ 7๐ฅ/2 โ 2xb = โ7๐ฅ/2 b = โ7๐ฅ/(2(โ2๐ฅ)) b = 7/4 Now, in our equation x2 โ7๐ฅ/2+3/2=0 Adding and subtracting (7/4)^2 x2 โ7๐ฅ/2+3/2+(7/4)^2โ(7/4)^2=0 x2 +(7/4)^2โ7๐ฅ/2+3/2โ(7/4)^2=0 (๐ฅโ 7/4)^2+3/2 โ(7/4)^2=0 (๐ฅโ 7/4)^2+3/2โ49/16=0 (๐ฅโ 7/4)^2+(3(8) โ 49)/16=0 (๐ฅโ 7/4)^2+(24 โ 49)/16=0 (๐ฅโ 7/4)^2โ25/16=0 (๐ฅโ7/4)^2=25/16 (๐ฅโ7/4)^2=(5/4)^2 Cancelling square both sides ๐ฅโ7/4 = ยฑ 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 โ 4 = 0 2x2 + x โ 4 = 0 Dividing whole equation by 2 (2๐ฅ2 + ๐ฅ โ 4)/2=0/2 2๐ฅ2/2+๐ฅ/2โ4/2=0 x2 + ๐ฅ/2โ2=0 We know that (a + b)2 = a2 + 2ab + b2 Here, a = x & 2ab = ๐ฅ/2 2xb = ๐ฅ/2 2b = 1/2 b = 1/2 ร 1/2 b = 1/4 Now, in our equation x2 + ๐ฅ/2โ2=0 Adding and subtracting (1/4)^2 x2 + ๐ฅ/2โ2+(1/4)^2โ(1/4)^2= 0 x2 + ๐ฅ/2+(1/4)^2โ 2 โ (1/4)^2=0 (๐ฅ+1/4)^2โ2โ(1/4)^2= 0 (๐ฅ+1/4)^2โ2 โ1/16=0 (๐ฅ+1/4)^2=2+1/16 (๐ฅ+1/4)^2=(2(16) + 1)/16 (๐ฅ+1/4)^2=(32 + 1)/16 (๐ฅ+1/4)^2=33/16 (๐ฅ+1/4)^2=(โ33/4)^2 Cancelling square both sides ๐ฅ+1/4 = ยฑ โ33/4 Solving So, the root of the equation are x = (โ33 โ 1)/4 & x = (โ(โ33 + 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 + 4โ3 ๐ฅ+3=0 4x2 + 4 โ3 ๐ฅ+3=0 Dividing whole equation 4 (4๐ฅ^2+ 4 โ3 ๐ฅ+ 3)/4=0/4 (4๐ฅ^2)/4 + (4 โ3)/4 x + 3/4=0 x2 + โ3 ๐ฅ+ 3/4 = 0 We know that (a + b)2 = a2 + 2ab + b2 Here, a = x & 2ab = โ3 ๐ฅ 2xb = โ3 ๐ฅ 2b = โ3 b = โ3/2 Now, in our equation x2 + โ3 ๐ฅ+3/4=0 Adding and subtracting (โ3/2)^2 x2 + โ3 ๐ฅ+3/4+(โ3/2)^2โ(โ3/2)^2=0 x2 + โ3 ๐ฅ+(โ3/2)^2+3/4โ(โ3/2)^2=0 (๐ฅ+โ3/2 )^2+3/4โ(โ3/2)^2=0 (๐ฅ+โ3/2 )^2+3/4โ3/4=0 (๐ฅ+โ3/2 )^2=0 (๐ฅ+โ3/2 )^2=02 Cancelling square both sides (๐ฅ+โ3/2 )^2= ยฑ 0 So, the root of the equation are x = (โโ3)/2 & x = (โโ3)/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 (2๐ฅ2 + ๐ฅ + 4)/2=0/2 2๐ฅ2/2+๐ฅ/2 + 4/2=0 x2 + ๐ฅ/2+2=0 We know that (a + b)2 = a2 + 2ab + b2 Here, a = x & 2ab = ๐ฅ/2 2xb = ๐ฅ/2 2b = 1/2 b = 1/2ร1/2 b = 1/4 Now, in our equation x2 + ๐ฅ/2+2=0 Adding and subtracting (1/4)^2 x2 + ๐ฅ/2 +2+(1/4)^2โ(1/4)^2=0 "x2 + " ๐ฅ/2+(1/4)^2+2โ(1/4)^2=0 (๐ฅ+1/4)^2+2โ(1/4)^2=0 (๐ฅ+1/4)^2+2โ1/16=0 (๐ฅ+1/4)^2+((32 โ 1)/16)=0 (๐ฅ+1/4)^2+ 31/16=0 (๐ฅ+1/4)^2=(โ31)/16 Since square of any number cannot be negative So, answer does not exist

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.