Question

Question 1 Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x^2 - 7x + 3 = 0
(ii) 2x^2 + x - 4 = 0
(iii) 4x^2 + 4(3x)^0.5 + 3 = 0
(iv) 2x^2 + x + 4 = 0

Class 10 - Math - Quadratic Equations Page 87

Answer 1

(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

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