Question

$\red{\sf 2x^{2} -7x + 3 = 0}$(solve by complete squaring method)

​

Answer 1

Answer:

this answer of your question

Answer 2

Step-by-step explanation:

Given : 2x² –7x + 3 = 0

On dividing the whole equation by 2,

(x² - 7x/2 + 3/2) = 0

Shift the constant term on RHS

x² - 7x/2 = - 3/2

Add square of the ½ of the coefficient of x on both sides

On adding (½ of 7/2)² = (7/4)² both sides

x² - 7x/2 + (7/4)²= - 3/2 + (7/4)²

Write the LHS in the form of perfect square

(x - 7/4)² = - 3/2 + 49/16

[a² - 2ab + b² = (a - b)²]

(x - 7/4)² = (-3 × 8 + 49)/16

(x - 7/4)² = (-24 + 49)/16

(x - 7/4)² = 25/16

On taking square root on both sides

(x - 7/4) = √(25/16)

(x - 7/4) = ± 5/4

On shifting constant term (-7/4) to RHS

x = ± 5/4 + 7/4

x = 5/4 + 7/4

[Taking +ve sign]

x = (5 +7)/4

x = 12/4

x = 3

x = - 5/4 + 7/4

[Taking -ve sign]

x = (- 5 + 7)/4

x = 2/4

x = 1/2

Hence, the roots of the given equation are 3 & ½.