Question

1..Find the root of the following quadratic equation, if they exist, by the method of completing the square :


$2x {}^{2} - 7x + 3 = 0$
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Answer 1

${(a + b)}^{2} = {a}^{2} + 2ab + {b}^{2} \\ 2 {x}^{2} - 7x + 3 = 0 \\ 2( {x}^{2} - \frac{7}{2} x) = - 3 \\ {x}^{2} - \frac{7}{2} x = - \frac{3}{2} \\ {x}^{2} - 2 \times \frac{7}{4} x = - \frac{3}{2} \\ adding \: { (\frac{7}{4} )}^{2} \: on \: both \: sides \\ {x}^{2} - 2 \times \frac{7}{4} x + {( \frac{7}{4} )}^{2} = - \frac{3}{2} + { (\frac{7}{4}) }^{2} \\ {x}^{2} - 2 \times \frac{7}{4} x + { (\frac{7}{4} )}^{2} = \frac{25}{16} = { (\frac{5}{4} )}^{2} \\ = {(x - \frac{7}{4} )}^{2} = { (\frac{5}{4} )}^{2} \\ = x - \frac{7}{4} = \frac{ |5| }{4} \\ x = \frac{7 + |5| }{4} \\ x = \frac{7 + 5}{4} = \frac{12}{4} = 3 \\ x = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2}$
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.

HOPE THIS ANSWER WILL HELP YOU...