Answer:
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.
Step-by-step explanation:
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