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Exercise 4.3 ka first question |
| Answer» 1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:(i) 2x2\xa0– 7x\xa0+3 = 0(ii)\xa02x2\xa0+\xa0x\xa0– 4 = 0(iii)\xa04x2\xa0+ 4√3x\xa0+ 3 = 0(iv)\xa02x2\xa0+\xa0x\xa0+ 4 = 0Solutions:(i) 2x2\xa0–\xa07x\xa0+ 3 = 0⇒ 2x2\xa0–\xa07x\xa0= – 3Dividing by 2 on both sides, we get⇒ x2\xa0-7x/2 = -3/2⇒ x2\xa0-2 × x ×7/4 = -3/2On adding (7/4)2\xa0to both sides of equation, we get⇒ (x)2-2×x×7/4 +(7/4)2\xa0= (7/4)2-3/2⇒ (x-7/4)2\xa0= (49/16) – (3/2)⇒(x-7/4)2\xa0= 25/16⇒(x-7/4)2\xa0= ±5/4⇒\xa0x\xa0= 7/4 ± 5/4⇒\xa0x\xa0= 7/4 + 5/4 or x = 7/4 – 5/4⇒ x = 12/4 or x = 2/4⇒\xa0x = 3 or x = 1/2(ii) 2x2\xa0+\xa0x\xa0– 4 = 0⇒ 2x2\xa0+\xa0x\xa0= 4Dividing both sides of the equation by 2, we get⇒\xa0x2\xa0+x/2 = 2Now on adding (1/4)2\xa0to both sides of the equation, we get,⇒ (x)2\xa0+\xa02 ×\xa0x\xa0× 1/4 + (1/4)2\xa0= 2\xa0+ (1/4)2⇒ (x\xa0+ 1/4)2\xa0= 33/16⇒\xa0x\xa0+ 1/4 = ± √33/4⇒\xa0x\xa0= ± √33/4 – 1/4⇒\xa0x\xa0= ± √33-1/4Therefore, either x\xa0= √33-1/4 or\xa0x\xa0= -√33-1/4(iii) 4x2\xa0+ 4√3x\xa0+ 3 = 0Converting the equation into a2+2ab+b2\xa0form, we get,⇒ (2x)2\xa0+ 2 × 2x\xa0× √3\xa0+ (√3)2\xa0= 0⇒ (2x\xa0+ √3)2\xa0= 0⇒ (2x\xa0+ √3) = 0 and (2x\xa0+ √3) = 0Therefore, either\xa0x\xa0= -√3/2 or\xa0x\xa0= -√3/2.(iv) 2x2\xa0+\xa0x\xa0+ 4 = 0⇒ 2x2\xa0+\xa0x\xa0= -4Dividing both sides of the equation by 2, we get⇒\xa0x2\xa0+ 1/2x\xa0= 2⇒\xa0x2\xa0+ 2\xa0×\xa0x\xa0× 1/4 = -2By adding (1/4)2\xa0to both sides of the equation, we get⇒ (x)2\xa0+\xa02 ×\xa0x\xa0× 1/4 + (1/4)2\xa0= (1/4)2\xa0– 2⇒ (x\xa0+ 1/4)2\xa0= 1/16 – 2⇒ (x\xa0+ 1/4)2\xa0= -31/16As we know, the square of numbers cannot be negative.Therefore, there is no real root for the given equation, 2x2\xa0+\xa0x\xa0+ 4 = 0. | |