Social Sciences, asked by Anonymous, 1 year ago

Find roots of the following quadratic equations , if they exist , by the metjod of completing the square
(1) 2x²- 7x + 3= 0
(2) 2x²+x - 4 = 0
(3) 4x²+ 4√3x + 3= 0
(4) 2x²+ x + 4 = 0

Answers

Answered by TrapNation
11

(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.
Answered by Hacker20
5
(i) 2x2 – 7x + 3 = 0

2x2 – 7x = - 3

x2 – 7x/2 = -3/2

x2 – 2 × x × 7/4 = -3/2

(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 = 25/1

(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

x2 + x/2 = 2

(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


x2 + 1/2x = 2

x2 + 2 × x × 1/4 = -2


(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
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