Question

find the roots of the equation 2x^2-7x+3=0 by completing the square method

Answer 1

2x^2-7x=-3
X^2-7x/2=-3/2
X^2-2x1/2x7x/2=-3/2
X^2-2x1/2x7x/2+(7/4)^2=-3/2+(7/4)^2
(X-7/4)^2=-3/2+49/16
(X-7/4)^2=25/16
X-7/4=+or -5/4
x=+5/4+7/4
X=12/4=3
Also
X=-5/4+7/4
2/4=1/2
Hence x=3 and 1/2

Answer 2

SOLUTION :  

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.

HOPE THIS ANSWER WILL HELP YOU...