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
Ask Download

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

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
Jail