Question

(ii) 2x2 + x-4=0
(iv) 2x2 + x +4=0​

Answer 1

Answer:

(ii) 4+x-4=0

4-4+x=0

x=0

(iv) 4+x+4=0

4+4+x=0

8+x=0

x=8

Answer 2

(ii) 2x2 + x – 4 = 0 ⇒ 2x2 + x = 4

Dividing both sides of the equation by 2, we get

⇒ x2 +x/2 = 2

Now on adding (1/4)2 to both sides 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 = √33-1/4 or x = -√33-1/4

(iv) 2x2 + x + 4 = 0 ⇒ 2x2 + x = -4

Dividing both sides of the equation by 2, we get

⇒ x2 + 1/2x = 2

⇒ x2 + 2 × x × 1/4 = -2

By adding (1/4)2 to both sides of 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 be negative.

Therefore, there is no real root for the given equation, 2x2 + x + 4 = 0.