Question

QI. Find the roots of the following quadratic equations
exist, by the
the method of completing the
y they exist
square
2x² - 7x+3=0​

Answer 1

$\huge\bf { \red S \green O \pink L \blue U \orange T \purple I \red O \pink N \green{...}}$

→ 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 and 1/2