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Ex 4.1, 1 - Check whether following are quadratic equations - Ex 4.1

  1. Chapter 4 Class 10 Quadratic Equations
  2. Serial order wise
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Ex 4.1 ,1 Check whether the following are quadratic equations : (i) (π‘₯+1)^2 = 2(x – 3) (π‘₯+1)^2 = 2(x – 3) π‘₯2+12+2Γ—π‘₯Γ—1=2(π‘₯βˆ’3) π‘₯2 + 1 + 2π‘₯ = 2π‘₯ – 6 π‘₯2 + 1+ 2π‘₯ – 2π‘₯ + 6 = 0 π‘₯2 + 1 +6 = 0 π‘₯2 + 7 = 0 π‘₯2 + 0π‘₯ + 7 = 0 Since , it is of the form π‘Žπ‘₯2 + 𝑏π‘₯ + 𝑐 =0 Where π‘Ž = 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) (π‘₯ – 2)(π‘₯ + 1)= (π‘₯ – 1)(π‘₯ + 3) π‘₯ (π‘₯ + 1) – 2 (π‘₯ + 1) = π‘₯ (π‘₯ + 3) – 1 (π‘₯ + 3) π‘₯2 + π‘₯ – 2π‘₯ – 2 = π‘₯2 + 3π‘₯ – π‘₯ – 3 π‘₯2 + π‘₯ – 2π‘₯ – 2 – π‘₯2 – 3π‘₯ + π‘₯ + 3 = 0 (π‘₯2 – π‘₯2 ) + (π‘₯ – 2π‘₯ – 3π‘₯ + π‘₯ ) – 2 + 3 = 0 0 – 3π‘₯ + 1 = 0 – 3π‘₯ + 1 = 0 Since , highest power is 1 not 2, It is not in the form of π‘Žπ‘₯2 + 𝑏π‘₯ + 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) (π‘₯ – 3) (2π‘₯ + 1) = π‘₯ (π‘₯ + 5) π‘₯ (2π‘₯ + 1) – 3(2π‘₯ + 1) = π‘₯ (π‘₯ + 5) 2π‘₯2 + π‘₯ – 6π‘₯ βˆ’3 = π‘₯2 + 5π‘₯ 2π‘₯2 + π‘₯ – 6π‘₯ – 3 – π‘₯2 – 5π‘₯ = 0 2π‘₯2 – π‘₯2 +π‘₯ – 6π‘₯ – 5π‘₯ – 3 = 0 π‘₯2 – 10 π‘₯ – 3 = 0 Since, the equation is of the form π‘Žπ‘₯2 + 𝑏π‘₯ + 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) (2π‘₯ – 1)(π‘₯ – 3)= (π‘₯ + 5)(π‘₯ – 1) 2π‘₯ (π‘₯ – 3) – 1 (π‘₯ – 3) = π‘₯ (π‘₯ – 1) + 5 (π‘₯ – 1) 2π‘₯2 – 6π‘₯ – π‘₯ + 3 = π‘₯2 – π‘₯ + 5π‘₯ – 5 2π‘₯2 – 6π‘₯ – π‘₯ + 3 – π‘₯2 + π‘₯ – 5π‘₯ + 5 = 0 2π‘₯2 – π‘₯2 – 6π‘₯ – π‘₯ + π‘₯ – 5π‘₯ + 3 + 5 = 0 π‘₯2 – 11π‘₯ + 8 = 0 Since it is of the form π‘Žπ‘₯2 + 𝑏π‘₯ + 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 π‘₯2 + 3π‘₯ + 1 = (π‘₯ – 2)^2 π‘₯2 + 3π‘₯ +1 = π‘₯2+22βˆ’2 Γ—π‘₯Γ—2 π‘₯2 +3π‘₯ +1 = π‘₯2 + 4 – 4π‘₯ π‘₯2 + 3π‘₯ + 1 – π‘₯2 – 4+ 4π‘₯= 0 π‘₯2 – π‘₯2 + 3π‘₯ + 4π‘₯ + 1 – 4 = 0 0 + 7π‘₯ – 3 = 0 7π‘₯ – 3 = 0 Since , highest power is 1 not 2, It is not in the form of π‘Žπ‘₯2 + 𝑏π‘₯ + 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 Γ—π‘₯Γ—2(π‘₯+2)=2π‘₯ (x2 – 1) π‘₯3 +8+6π‘₯(π‘₯+2)=2π‘₯3βˆ’2π‘₯ π‘₯2 + 8 + 6π‘₯2 + 12π‘₯ = 2π‘₯3 – 2π‘₯ π‘₯3 + 8 + 6π‘₯2 + 12π‘₯ – 2π‘₯3 + 2π‘₯ = 0 π‘₯3 – 2π‘₯3 + 6π‘₯2 + 12π‘₯ + 2π‘₯ + 8 = 0 – π‘₯3 + 6π‘₯2 +14π‘₯ + 8 = 0 Since highest power is 3 and not 2, It is not in the form of π‘Žπ‘₯2 + 𝑏π‘₯ + 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 π‘₯3 – 4π‘₯2 – π‘₯+ 1 = (π‘₯ – 2)^3 π‘₯3 – 4π‘₯2 – π‘₯+1 = π‘₯3 –23 βˆ’3Γ—π‘₯Γ—2(xβˆ’2) π‘₯3 – 4π‘₯2 – π‘₯+1 = π‘₯3 – 8 – 6π‘₯ (π‘₯ – 2) π‘₯3 – 4π‘₯2 – π‘₯ +1 = π‘₯3 – 8 – 6π‘₯2 + 12 π‘₯ π‘₯3 – 4π‘₯2 – π‘₯ + 1 – π‘₯3 + 8 + 6π‘₯2 – 12π‘₯ = 0 π‘₯3 – π‘₯3 – 4π‘₯2 + 6π‘₯2 – π‘₯ – 12π‘₯ + 1 + 8 = 0 0 + 2π‘₯2 – 13π‘₯ + 9 = 0 2π‘₯2 – 13π‘₯ + 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

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