Question

Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x2 – 7x + 3 = 0
(ii) 2x2 + x – 4 = 0
(iii) 4x2 + 4√3x + 3 = 0
(iv) 2x2 + x + 4 = 0


oyr rv me jeenu hu bio pe baat kyu ni kr rha -_-​

Answer 1

Answer:

(i) 2x2 – 7x + 3 = 0

⇒ 2x2 – 7x = - 3

On dividing both sides of the equation by 2, we get

⇒ x2 – 7x/2  = -3/2

⇒ x2 – 2 × x ×  7/4 = -3/2

On adding (7/4)2 to both sides of 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) = ± 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 1/2

(ii) 2x2 + x – 4 = 0

⇒ 2x2 + x = 4

On dividing both sides of the equation, we get

⇒ x2 + x/2 = 2

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

⇒ x = √33-1/4 or x = -√33-1/4

(iii) 4x2 + 4√3x + 3 = 0

⇒ (2x)2 + 2 × 2x × √3 + (√3)2 = 0

⇒ (2x + √3)2 = 0

⇒ (2x + √3) = 0 and (2x + √3) = 0

⇒ x = -√3/2 or x = -√3/2

(iv) 2x2 + x + 4 = 0

⇒ 2x2 + x = -4

On dividing both sides of the equation, we get

⇒ x2 + 1/2x = 2

⇒ x2 + 2 × x × 1/4 = -2

On 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

However, the square of number cannot be negative.

Therefore, there is no real root for the given equation.

Answer 2

Janta hu aap Jeena ho actually main bohot kaam on ata hu isliye tumse baat nhi ho pata hain

Btw tumne mera name ka spelling wrong likha hain Mera name ka spelling Aryan hain