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Chapter 4 Class 10 Quadratic Equations

Serial order wise

Last updated at March 22, 2023 by Teachoo

Ex 4.1 ,1 Check whether the following are quadratic equations : (i) (𝑥+1)^2 = 2(x – 3) (𝑥+1)^2 = 2(x – 3) 𝑥2+12+2×𝑥×1=2(𝑥−3) 𝑥2 + 1 + 2𝑥 = 2𝑥 – 6 𝑥2 + 1+ 2𝑥 – 2𝑥 + 6 = 0 𝑥2 + 1 +6 = 0 𝑥2 + 7 = 0 𝑥2 + 0𝑥 + 7 = 0 Since , it is of the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 =0 Where 𝑎 = 1, 𝑏 = 0, 𝑐 = 7 Hence, it is a quadratic equation Ex 4.1 ,1 Check whether the following are quadratic equations : (ii) x2 – 2x = (-2) (3 – x) x2 – 2x = (-2) (3 – x) x2 – 2x = (-2)3 – (–2)x x2 – 2x = –6 + 2x x2 – 2x – 2x + 6 = 0 x2 – 4x + 6 = 0 It is the form of ax2 + bx + c = 0 Where a = 1, b = – 4 , c = 6 Hence, it is a quadratic equation . Ex 4.1 ,1 Check whether the following are quadratic equations : (iii) (x – 2)(x + 1) = (x – 1)(x + 3) (𝑥 – 2)(𝑥 + 1)= (𝑥 – 1)(𝑥 + 3) 𝑥 (𝑥 + 1) – 2 (𝑥 + 1) = 𝑥 (𝑥 + 3) – 1 (𝑥 + 3) 𝑥2 + 𝑥 – 2𝑥 – 2 = 𝑥2 + 3𝑥 – 𝑥 – 3 𝑥2 + 𝑥 – 2𝑥 – 2 – 𝑥2 – 3𝑥 + 𝑥 + 3 = 0 (𝑥2 – 𝑥2 ) + (𝑥 – 2𝑥 – 3𝑥 + 𝑥 ) – 2 + 3 = 0 0 – 3𝑥 + 1 = 0 – 3𝑥 + 1 = 0 Since , highest power is 1 not 2, It is not in the form of 𝑎𝑥2 + 𝑏𝑥 + 𝑐 =0 Hence, it is not a quadratic equation. Ex 4.1 ,1 Check whether the following are quadratic equations : (iv) (x – 3)(2x +1) = x(x + 5) (𝑥 – 3) (2𝑥 + 1) = 𝑥 (𝑥 + 5) 𝑥 (2𝑥 + 1) – 3(2𝑥 + 1) = 𝑥 (𝑥 + 5) 2𝑥2 + 𝑥 – 6𝑥 −3 = 𝑥2 + 5𝑥 2𝑥2 + 𝑥 – 6𝑥 – 3 – 𝑥2 – 5𝑥 = 0 2𝑥2 – 𝑥2 +𝑥 – 6𝑥 – 5𝑥 – 3 = 0 𝑥2 – 10 𝑥 – 3 = 0 Since, the equation is of the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 = 0 Where, a = 1, b = – 10, c = – 3 Hence it is a quadratic equation. Ex 4.1 ,1 Check whether the following are quadratic equations : (v) (2x – 1)(x – 3) = (x + 5)(x – 1) (2𝑥 – 1)(𝑥 – 3)= (𝑥 + 5)(𝑥 – 1) 2𝑥 (𝑥 – 3) – 1 (𝑥 – 3) = 𝑥 (𝑥 – 1) + 5 (𝑥 – 1) 2𝑥2 – 6𝑥 – 𝑥 + 3 = 𝑥2 – 𝑥 + 5𝑥 – 5 2𝑥2 – 6𝑥 – 𝑥 + 3 – 𝑥2 + 𝑥 – 5𝑥 + 5 = 0 2𝑥2 – 𝑥2 – 6𝑥 – 𝑥 + 𝑥 – 5𝑥 + 3 + 5 = 0 𝑥2 – 11𝑥 + 8 = 0 Since it is of the form 𝑎𝑥2 + 𝑏𝑥 + 𝑐 =0 Where a = 1, b = – 11, c = 8 Hence it is a quadratic equation. Ex 4.1 ,1 Check whether the following are quadratic equations : (vi) x2 + 3x + 1 = (x – 2)2 𝑥2 + 3𝑥 + 1 = (𝑥 – 2)^2 𝑥2 + 3𝑥 +1 = 𝑥2+22−2 ×𝑥×2 𝑥2 +3𝑥 +1 = 𝑥2 + 4 – 4𝑥 𝑥2 + 3𝑥 + 1 – 𝑥2 – 4+ 4𝑥= 0 𝑥2 – 𝑥2 + 3𝑥 + 4𝑥 + 1 – 4 = 0 0 + 7𝑥 – 3 = 0 7𝑥 – 3 = 0 Since , highest power is 1 not 2, It is not in the form of 𝑎𝑥2 + 𝑏𝑥 + 𝑐 =0 Hence, it is not a quadratic equation. Ex 4.1 ,1 Check whether the following are quadratic equations : (vii) (x + 2)3 = 2x (x2 – 1) (x + 2)3 = 2x (x2 – 1) x3 + 23 + 3 ×𝑥×2(𝑥+2)=2𝑥 (x2 – 1) 𝑥3 +8+6𝑥(𝑥+2)=2𝑥3−2𝑥 𝑥2 + 8 + 6𝑥2 + 12𝑥 = 2𝑥3 – 2𝑥 𝑥3 + 8 + 6𝑥2 + 12𝑥 – 2𝑥3 + 2𝑥 = 0 𝑥3 – 2𝑥3 + 6𝑥2 + 12𝑥 + 2𝑥 + 8 = 0 – 𝑥3 + 6𝑥2 +14𝑥 + 8 = 0 Since highest power is 3 and not 2, It is not in the form of 𝑎𝑥2 + 𝑏𝑥 + 𝑐 =0 Hence, it is not a quadratic equation Ex 4.1 ,1 Check whether the following are quadratic equations : (viii) x3 – 4x2 – x + 1 = (x – 2)3 𝑥3 – 4𝑥2 – 𝑥+ 1 = (𝑥 – 2)^3 𝑥3 – 4𝑥2 – 𝑥+1 = 𝑥3 –23 −3×𝑥×2(x−2) 𝑥3 – 4𝑥2 – 𝑥+1 = 𝑥3 – 8 – 6𝑥 (𝑥 – 2) 𝑥3 – 4𝑥2 – 𝑥 +1 = 𝑥3 – 8 – 6𝑥2 + 12 𝑥 𝑥3 – 4𝑥2 – 𝑥 + 1 – 𝑥3 + 8 + 6𝑥2 – 12𝑥 = 0 𝑥3 – 𝑥3 – 4𝑥2 + 6𝑥2 – 𝑥 – 12𝑥 + 1 + 8 = 0 0 + 2𝑥2 – 13𝑥 + 9 = 0 2𝑥2 – 13𝑥 + 9 = 0 It is of the form ax2 + bx + c = 0 Where a = 2, b = – 13 and c = 9 Hence, it is a quadratic equation