Ex 4.3, 1 - Find roots by completing the square (i) 2x2 - - Solving by completing square

  1. Chapter 4 Class 10 Quadratic Equations
  2. Serial order wise

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 โ€“ 7x + 3 = 0 2x2 โ€“ 7x +3 = 0 Dividing by 2 (2๐‘ฅ2 โˆ’ 7๐‘ฅ + 3 = 0)/2=0/2 2๐‘ฅ2/2 โ€“ 7๐‘ฅ/2 + 3/2=0 x2 โ€“ 7๐‘ฅ/2+3/2=0 We know that (a โ€“ b)2 = a2 โ€“ 2ab + b2 Here, a = x & โ€“ 2ab = โ€“ 7๐‘ฅ/2 โ€“ 2xb = โˆ’7๐‘ฅ/2 b = โˆ’7๐‘ฅ/(2(โˆ’2๐‘ฅ)) b = 7/4 Now, in our equation x2 โˆ’7๐‘ฅ/2+3/2=0 Adding and subtracting (7/4)^2 x2 โˆ’7๐‘ฅ/2+3/2+(7/4)^2โˆ’(7/4)^2=0 x2 +(7/4)^2โˆ’7๐‘ฅ/2+3/2โˆ’(7/4)^2=0 (๐‘ฅโˆ’ 7/4)^2+3/2 โˆ’(7/4)^2=0 (๐‘ฅโˆ’ 7/4)^2+3/2โˆ’49/16=0 (๐‘ฅโˆ’ 7/4)^2+(3(8) โˆ’ 49)/16=0 (๐‘ฅโˆ’ 7/4)^2+(24 โˆ’ 49)/16=0 (๐‘ฅโˆ’ 7/4)^2โˆ’25/16=0 (๐‘ฅโˆ’7/4)^2=25/16 (๐‘ฅโˆ’7/4)^2=(5/4)^2 Cancelling square both sides ๐‘ฅโˆ’7/4 = 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 (2๐‘ฅ2 + ๐‘ฅ โˆ’ 4)/2=0/2 2๐‘ฅ2/2+๐‘ฅ/2โˆ’4/2=0 x2 + ๐‘ฅ/2โˆ’2=0 We know that (a + b)2 = a2 + 2ab + b2 Here, a = x & 2ab = ๐‘ฅ/2 2xb = ๐‘ฅ/2 2b = 1/2 b = 1/2 ร— 1/2 b = 1/4 Now, in our equation x2 + ๐‘ฅ/2โˆ’2=0 Adding and subtracting (1/4)^2 x2 + ๐‘ฅ/2โˆ’2+(1/4)^2โˆ’(1/4)^2= 0 x2 + ๐‘ฅ/2+(1/4)^2โ€“ 2 โ€“ (1/4)^2=0 (๐‘ฅ+1/4)^2โˆ’2โˆ’(1/4)^2= 0 (๐‘ฅ+1/4)^2โˆ’2 โˆ’1/16=0 (๐‘ฅ+1/4)^2=2+1/16 (๐‘ฅ+1/4)^2=(2(16) + 1)/16 (๐‘ฅ+1/4)^2=(32 + 1)/16 (๐‘ฅ+1/4)^2=33/16 (๐‘ฅ+1/4)^2=(โˆš33/4)^2 Cancelling square both sides ๐‘ฅ+1/4 = ยฑ โˆš33/4 Solving So, the root of the equation are x = (โˆš33 โˆ’ 1)/4 & x = (โˆ’(โˆš33 + 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 + 4โˆš3 ๐‘ฅ+3=0 4x2 + 4 โˆš3 ๐‘ฅ+3=0 Dividing whole equation 4 (4๐‘ฅ^2+ 4 โˆš3 ๐‘ฅ+ 3)/4=0/4 (4๐‘ฅ^2)/4 + (4 โˆš3)/4 x + 3/4=0 x2 + โˆš3 ๐‘ฅ+ 3/4 = 0 We know that (a + b)2 = a2 + 2ab + b2 Here, a = x & 2ab = โˆš3 ๐‘ฅ 2xb = โˆš3 ๐‘ฅ 2b = โˆš3 b = โˆš3/2 Now, in our equation x2 + โˆš3 ๐‘ฅ+3/4=0 Adding and subtracting (โˆš3/2)^2 x2 + โˆš3 ๐‘ฅ+3/4+(โˆš3/2)^2โˆ’(โˆš3/2)^2=0 x2 + โˆš3 ๐‘ฅ+(โˆš3/2)^2+3/4โˆ’(โˆš3/2)^2=0 (๐‘ฅ+โˆš3/2 )^2+3/4โˆ’(โˆš3/2)^2=0 (๐‘ฅ+โˆš3/2 )^2+3/4โˆ’3/4=0 (๐‘ฅ+โˆš3/2 )^2=0 (๐‘ฅ+โˆš3/2 )^2=02 Cancelling square both sides (๐‘ฅ+โˆš3/2 )^2= ยฑ 0 So, the root of the equation are x = (โˆ’โˆš3)/2 & x = (โˆ’โˆš3)/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 (2๐‘ฅ2 + ๐‘ฅ + 4)/2=0/2 2๐‘ฅ2/2+๐‘ฅ/2 + 4/2=0 x2 + ๐‘ฅ/2+2=0 We know that (a + b)2 = a2 + 2ab + b2 Here, a = x & 2ab = ๐‘ฅ/2 2xb = ๐‘ฅ/2 2b = 1/2 b = 1/2ร—1/2 b = 1/4 Now, in our equation x2 + ๐‘ฅ/2+2=0 Adding and subtracting (1/4)^2 x2 + ๐‘ฅ/2 +2+(1/4)^2โˆ’(1/4)^2=0 "x2 + " ๐‘ฅ/2+(1/4)^2+2โˆ’(1/4)^2=0 (๐‘ฅ+1/4)^2+2โˆ’(1/4)^2=0 (๐‘ฅ+1/4)^2+2โˆ’1/16=0 (๐‘ฅ+1/4)^2+((32 โˆ’ 1)/16)=0 (๐‘ฅ+1/4)^2+ 31/16=0 (๐‘ฅ+1/4)^2=(โˆ’31)/16 Since square of any number cannot be negative So, answer does not exist

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.