Last updated at Dec. 8, 2016 by Teachoo

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Ex 4.1 ,1 Check whether the following are quadratic equations : (i) (π₯+1)^2 = 2(x β 3) (π₯+1)^2 = 2(x β 3) π₯2+12+2Γπ₯Γ1=2(π₯β3) π₯2 + 1 + 2π₯ = 2π₯ β 6 π₯2 + 1+ 2π₯ β 2π₯ + 6 = 0 π₯2 + 1 +6 = 0 π₯2 + 7 = 0 π₯2 + 0π₯ + 7 = 0 Since , it is of the form ππ₯2 + ππ₯ + π =0 Where π = 1, π = 0, π = 7 Hence, it is a quadratic equation Ex 4.1 ,1 Check whether the following are quadratic equations : (ii) x2 β 2x = (-2) (3 β x) x2 β 2x = (-2) (3 β x) x2 β 2x = (-2)3 β (β2)x x2 β 2x = β6 + 2x x2 β 2x β 2x + 6 = 0 x2 β 4x + 6 = 0 It is the form of ax2 + bx + c = 0 Where a = 1, b = β 4 , c = 6 Hence, it is a quadratic equation . Ex 4.1 ,1 Check whether the following are quadratic equations : (iii) (x β 2)(x + 1) = (x β 1)(x + 3) (π₯ β 2)(π₯ + 1)= (π₯ β 1)(π₯ + 3) π₯ (π₯ + 1) β 2 (π₯ + 1) = π₯ (π₯ + 3) β 1 (π₯ + 3) π₯2 + π₯ β 2π₯ β 2 = π₯2 + 3π₯ β π₯ β 3 π₯2 + π₯ β 2π₯ β 2 β π₯2 β 3π₯ + π₯ + 3 = 0 (π₯2 β π₯2 ) + (π₯ β 2π₯ β 3π₯ + π₯ ) β 2 + 3 = 0 0 β 3π₯ + 1 = 0 β 3π₯ + 1 = 0 Since , highest power is 1 not 2, It is not in the form of ππ₯2 + ππ₯ + π =0 Hence, it is not a quadratic equation. Ex 4.1 ,1 Check whether the following are quadratic equations : (iv) (x β 3)(2x +1) = x(x + 5) (π₯ β 3) (2π₯ + 1) = π₯ (π₯ + 5) π₯ (2π₯ + 1) β 3(2π₯ + 1) = π₯ (π₯ + 5) 2π₯2 + π₯ β 6π₯ β3 = π₯2 + 5π₯ 2π₯2 + π₯ β 6π₯ β 3 β π₯2 β 5π₯ = 0 2π₯2 β π₯2 +π₯ β 6π₯ β 5π₯ β 3 = 0 π₯2 β 10 π₯ β 3 = 0 Since, the equation is of the form ππ₯2 + ππ₯ + π = 0 Where, a = 1, b = β 10, c = β 3 Hence it is a quadratic equation. Ex 4.1 ,1 Check whether the following are quadratic equations : (v) (2x β 1)(x β 3) = (x + 5)(x β 1) (2π₯ β 1)(π₯ β 3)= (π₯ + 5)(π₯ β 1) 2π₯ (π₯ β 3) β 1 (π₯ β 3) = π₯ (π₯ β 1) + 5 (π₯ β 1) 2π₯2 β 6π₯ β π₯ + 3 = π₯2 β π₯ + 5π₯ β 5 2π₯2 β 6π₯ β π₯ + 3 β π₯2 + π₯ β 5π₯ + 5 = 0 2π₯2 β π₯2 β 6π₯ β π₯ + π₯ β 5π₯ + 3 + 5 = 0 π₯2 β 11π₯ + 8 = 0 Since it is of the form ππ₯2 + ππ₯ + π =0 Where a = 1, b = β 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 β 2)2 π₯2 + 3π₯ + 1 = (π₯ β 2)^2 π₯2 + 3π₯ +1 = π₯2+22β2 Γπ₯Γ2 π₯2 +3π₯ +1 = π₯2 + 4 β 4π₯ π₯2 + 3π₯ + 1 β π₯2 β 4+ 4π₯= 0 π₯2 β π₯2 + 3π₯ + 4π₯ + 1 β 4 = 0 0 + 7π₯ β 3 = 0 7π₯ β 3 = 0 Since , highest power is 1 not 2, It is not in the form of ππ₯2 + ππ₯ + π =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 β 1) (x + 2)3 = 2x (x2 β 1) x3 + 23 + 3 Γπ₯Γ2(π₯+2)=2π₯ (x2 β 1) π₯3 +8+6π₯(π₯+2)=2π₯3β2π₯ π₯2 + 8 + 6π₯2 + 12π₯ = 2π₯3 β 2π₯ π₯3 + 8 + 6π₯2 + 12π₯ β 2π₯3 + 2π₯ = 0 π₯3 β 2π₯3 + 6π₯2 + 12π₯ + 2π₯ + 8 = 0 β π₯3 + 6π₯2 +14π₯ + 8 = 0 Since highest power is 3 and not 2, It is not in the form of ππ₯2 + ππ₯ + π =0 Hence, it is not a quadratic equation Ex 4.1 ,1 Check whether the following are quadratic equations : (viii) x3 β 4x2 β x + 1 = (x β 2)3 π₯3 β 4π₯2 β π₯+ 1 = (π₯ β 2)^3 π₯3 β 4π₯2 β π₯+1 = π₯3 β23 β3Γπ₯Γ2(xβ2) π₯3 β 4π₯2 β π₯+1 = π₯3 β 8 β 6π₯ (π₯ β 2) π₯3 β 4π₯2 β π₯ +1 = π₯3 β 8 β 6π₯2 + 12 π₯ π₯3 β 4π₯2 β π₯ + 1 β π₯3 + 8 + 6π₯2 β 12π₯ = 0 π₯3 β π₯3 β 4π₯2 + 6π₯2 β π₯ β 12π₯ + 1 + 8 = 0 0 + 2π₯2 β 13π₯ + 9 = 0 2π₯2 β 13π₯ + 9 = 0 It is of the form ax2 + bx + c = 0 Where a = 2, b = β 13 and c = 9 Hence, it is a quadratic equation

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CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .