Question

what is roots of quadratic equation 2x^2-7x+3=0. solve by method of completing square.

Answer 1

$2 {x}^{2} - 7x + 3 = 0 \\ = > 2( {x}^{2} - \frac{7x}{2} + \frac{3}{2} ) = 0 \\ = > 2( {x}^{2} - 2 ( \frac{7}{4} )x + \frac{49}{16} ) = \frac{49}{8} - 3 \\ = > ( {x - \frac{7}{4}) }^{2} = ({ \frac{5}{4} )}^{2} \\ = > (x - \frac{7}{4} ) = | \frac{5}{4} | \\ = > x = \frac{5}{4} + \frac{7}{4} \\ = > x = - \frac{5}{4} + \frac{7}{4} \\ = > x = 3 \\ = > x = 1$
Hope it helps

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.

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