This page samacheer kalvi math solution for exercise 3.15 part1 is going to provide you solution for every problems that you find in the exercise no 3.15
(1) Solve the following quadratic equations by completing the square.
(i) x² + 6 x - 7 = 0
Solution:
x²+ 2 x x x 3 + 3² - 3² - 7 = 0
(x + 3)² - 9 - 7 = 0
(x + 3)² - 16 = 0
(x + 3)² = 16
x + 3 = √ 16
x + 3 = ± 4
x + 3 = 4 x + 3 = - 4
x = 4 - 3 x = - 4 - 3
x = 1 x = - 7
(ii) x² + 3 x + 1 = 0
Solution:
x² + 3 x + 1 = 0
x²+ 2 x x x (3/2) + (3/2)² - (3/2)² + 1 = 0
(x + (3/2))² - (9/4) + 1 = 0
(x + (3/2))² = (9/4) - 1
(x + (3/2))² = (5/4)
(x + (3/2)) = √(5/4)
x + (3/2) = ± (√5/2)
x + (3/2) = (√5/2) x + (3/2) = -(√5/2)
x = (√5/2) - (3/2) x = -(√5/2) - (3/2)
x = (√5 - 3)/2 x = (-√5 - 3)/2
In the page samacheer kalvi math solution for exercise 3.15 part1 we are going to see the solution of next problem
(iii) 2 x² + 5 x - 3 = 0
Solution:
2 x² + 5 x - 3 = 0
divide the whole equation by 2
x² + (5/2) x - (3/2) = 0
x² + 2 (5/2) x - (3/2) = 0
x² + 2 x (5/2) + (5/2)² - (5/2)²- (3/2) = 0
(x + (5/2))² - (25/4)- (3/2) = 0
(x + (5/2))² = (25/4) + (3/2)
(x + (5/2))² = (25 + 6)/4
(x + (5/2))² = 31/4
x + (5/2) = √(31/4)
x + (5/2) = ± √31/2
x + (5/2) = √31/2 x + (5/2) = -√31/2
x = (√31/2) - (5/2) x = (-√31/2) - (5/2)
x = (√31 - 5)/2 x = (- √31-5)/2
(iv) 4 x² + 4 b x - (a² - b²) = 0
dividing the whole equation by 4,we get
x² + b x - (a² - b²)/4= 0
x = (a - b)/2 or x = (-a -b)/2
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