Ssamacheer Kalvi Math Solution for Exercise 3.12 Part 1

This page samacheer kalvi math solution for exercise 3.12 part 1 is going to provide you solution for every problems that you find in the exercise no 3.12

Samacheer Kalvi Math Solution for Exercise 3.12 part 1

(1) Find the square root of the following

(i) 196 a⁶ b⁸ c¹⁰

Solution:

          = √(196 a⁶ b⁸ c¹⁰ )

          = √(14 x 14 a³ x a³ x b⁴ x b⁴ x c⁵ x c⁵)

           = 14  a³bc⁵



(ii) 289 (a - b)⁴ (b - c)

Solution:

          = √289 (a - b)⁴ (b - c)

          = √(17 x 17 (a - b)² (a - b)² (b - c)³(b - c)³

           = 17 (a - b)² (b - c)³


In the page samacheer kalvi math solution for exercise 3.12 part 1  we are going to see the solution of next problem

(iii) (x + 11)² - 44 x

Solution:

(a + b)² = a² + 2 a b + b²

            = (x² + 2 x (11) + 11² - 44 x)

            = (x² + 22 x + 121 - 44 x)

            = (x² - 22 x + 121 )

           = (x - 11)²

           = x - 11


(iv) (x - y)² + 4 x y

Solution:

 (a - b)² = a² - 2 a b + b²

             = [(x - y)² + 4 x y]

             = [x² - 2 x y + y² + 4 x y]

             = (x² + 2 x y + y²)

             =
((x + y)²

             = (x + y)


(v) 121 x⁸ y÷ 81 x⁴ y

Solution:

            = √(121 x⁸ y⁶/ 81 x⁴ y⁸)

            = √(11 x 11 x xx xx y³x y³/ 9 x x² x x² x y x y⁴)

            = (11/9) ( x⁴  y³/ x² y)

            = (11/9) ( x² /y)


(vi) [64 (a + b)⁴(x - y)⁸(b - c)⁶]/[25 (x+ y)⁴ (a - b)⁶(b + c)¹⁰]

Solution:

    = [64 (a + b)⁴(x - y)⁸(b - c)⁶]/[25 (x+ y)⁴ (a - b)⁶(b + c)¹⁰]

   = (8/5) [(a + b)² (x - y)(b - c)³/(x+ y) ² (a - b)³(b + c)⁵]




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