Find two integers whose sum is -12 A. 12,6 B. -12,0 C. 13,1 D. -13,-1
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Check whether the following are quadratic equations:
(i) (x + 1)2 = 2(x – 3)(ii) x2 – 2x = (–2)(3 – x) (iii) (x – 2)(x + 1) = (x – 1)(x + 3)(iv) (x – 3)(2x + 1) = x(x + 5) (v) (2x – 1) (x – 3) – (x + 5) (x – 1)(vi) x2 + 3x +1 = (x – 2)2 (vii) (x + 2)3 = 2x(x2 – 1)(viii) x3 – 4x2 – × + 1 = (x – 2)3
Sol. (i) (x + 1)2 = 2(x – 3) We have: (x + 1)2 = 2 (x – 3) x2 + 2x + 1 = 2x – 6 ⇒ x2 + 2x + 1 – 2x + 6 = 0 ⇒ x2 + 70 Since x2 + 7 is a quadratic polynomial ∴ (x + 1)2 = 2(x – 3) is a quadratic equation. (ii) x2– 2x = (–2) (3 – x) We have: x2 – 2x = (– 2) (3 – x) ⇒ x2 – 2x = –6 + 2x ⇒ x2 – 2x – 2x + 6 = 0 ⇒ x2 – 4x + 6 = 0 Since x2 – 4x + 6 is a quadratic polynomial ∴ x2 – 2x = (–2) (3 – x) is a quadratic equation. (iii) (x – 2) (x + 1) = (x – 1) (x + 3) We have: (x – 2) (x + 1) = (x – 1) (x + 3) ⇒ x2 – x – 2 = x2 + 2x – 3 ⇒ x2 – x – 2 – x2 – 2x + 3 = 0 ⇒ –3x + 1 = 0 Since –3x + 1 is a linear polynomial ∴ (x – 2) (x + 1) = (x – 1) (x + 3) is not quadratic equation. (iv) (x – 3) (2x + 1) = x(x + 5) We have: (x – 3) (2x + 1) = x(x + 5) ⇒ 2x2 + x – 6x – 3 = x2 + 5x ⇒ 2x2 – 5x – 3 – x2 – 5x – 0 ⇒ x2 + 10x – 3 = 0 Since x2 + 10x – 3 is a quadratic polynomial ∴ (x – 3) (2x + 1) = x(x + 5) is a quadratic equation. (v) (2x – 1) (x – 3) = (x + 5) (x – 1) We have: (2x – 1) (x – 3) = (x + 5) (x – 1) ⇒ 2x2 – 6x – x + 3 = x2 – x + 5x – 5 ⇒ 2x2 – x2 – 6x – x + x – 5x + 3 + 5 = 0 ⇒ x2 – 11x + 8 = 0 Since x2 – 11x + 8 is a quadratic polynomial ∴ (2x – 1) (x – 3) = (x + 5) (x – 1) is a quadratic equation. (vi) x2 + 3x + 1 = (x – 2)2 We have: x2 + 3x + 1 = (x – 2)2 ⇒ x2 + 3x + 1 = x2 – 4x + 4 ⇒ x2 + 3x + 1 – x2 + 4x – 4 =0 ⇒ 7x – 3 = 0 Since 7x – 3 is a linear polynomial. ∴ x2 + 3x + 1 = (x – 2)2 is not a quadratic equation. (vii) (x + 2)3 = 2x(x2 – 1) We have: (x + 2)3 = 2x(x2 – 1) x3 + 3x2(2) + 3x(2)2 + (2)3 = 2x3 – 2x ⇒ x3 + 6x2 + 12x + 8 = 2x3 – 2x ⇒ x3 + 6x2 + 12x + 8 – 2x3 + 2x = 0 ⇒ –x3 + 6x2 + 14x + 8 = 0 Since –x3 + 6x2 + 14x + 8 is a polynomial of degree 3 ∴ (x + 2)3 = 2x(x2 – 1) is not a quadratic equation. (viii) x3 – 4x2 – x + 1 = (x – 2)3 We have: x3 – 4x2 – x + 1 = (x – 2)3 ⇒ x3 – 4x2 – x + 1 = x3 + 3x2(– 2) + 3x(– 2)2 + (– 2)3 ⇒ x3 – 4x2 – x + 1 = x3 – 6x2 + 12x – 8 ⇒ x3 – 4x2 – x – 1 – x3 + 6x2 – 12x + 8 = 0 2x2 – 13x + 9 = 0 Since 2x2 – 13x + 9 is a quadratic polynomial ∴ x3 – 4x2 – x + 1 = (x – 2)3 is a quadratic e
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