Question -
Answer -
(i) 2x2 –7x +3 = 0
(ii) 2x2 + x –4 = 0
(iii) 4x2 + 4√3x + 3 = 0
(iv) 2x2 + x +4 = 0
Solutions:
(i) 2x2 – 7x +3 = 0
⇒ 2x2 – 7x = – 3
Dividing by 2 on both sides, we get
⇒ x2 -7x/2 = -3/2
⇒ x2 -2 × x ×7/4 = -3/2
On adding (7/4)2 to both sidesof equation, we get
⇒ (x)2-2×x×7/4 +(7/4)2 = (7/4)2-3/2
⇒ (x-7/4)2 = (49/16) – (3/2)
⇒(x-7/4)2 = 25/16
⇒(x-7/4)2 = ±5/4
⇒ x = 7/4 ± 5/4
⇒ x = 7/4 + 5/4 or x = 7/4 – 5/4
⇒ x = 12/4 or x = 2/4
⇒ x = 3 or x = 1/2
(ii) 2x2 + x –4 = 0
⇒ 2x2 + x = 4
Dividing both sides of the equation by 2, weget
⇒ x2 +x/2 = 2
Now on adding (1/4)2 to bothsides of the equation, we get,
⇒ (x)2 + 2 × x × 1/4 + (1/4)2 =2 + (1/4)2
⇒ (x + 1/4)2 = 33/16
⇒ x + 1/4 = ± √33/4
⇒ x = ± √33/4 – 1/4
⇒ x = ± √33-1/4
Therefore, either x = √33-1/4or x = -√33-1/4
(iii) 4x2 +4√3x + 3 = 0
Converting the equation into a2+2ab+b2 form,we get,
⇒ (2x)2 + 2 × 2x × √3 + (√3)2 =0
⇒ (2x + √3)2 = 0
⇒ (2x + √3) = 0 and (2x + √3) = 0
Therefore, either x = -√3/2or x = -√3/2.
(iv) 2x2 + x +4 = 0
⇒ 2x2 + x = -4
Dividing both sides of the equation by 2, weget
⇒ x2 + 1/2x = 2
⇒ x2 + 2 × x × 1/4 = -2
By adding (1/4)2 to both sidesof the equation, we get
⇒ (x)2 + 2 × x × 1/4 + (1/4)2 =(1/4)2 – 2
⇒ (x + 1/4)2 = 1/16 – 2
⇒ (x + 1/4)2 = -31/16
As we know, the square of numbers cannot benegative.
Therefore, there is no real root for the givenequation, 2x2 + x + 4 = 0.