Ssamacheer Kalvi Math Solution for Exercise 3.12 Part 2

This page samacheer kalvi math solution for exercise 3.12 part 2 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 2

(2) Find the square root of the following

(i) 16 x² - 24 x + 9

Solution:

          = √(16 x² - 24 x + 9)

          = √(4 x)² - 2 (4x) (3) + 3²

          = √(4 x - 3)²

          = (4 x - 3)


(ii) (x² - 25)(x² + 8 x + 15)(x² - 2 x - 15)

Solution:

          = √(x² - 25)(x² + 8 x + 15)(x² - 2 x - 15)

          = √(x + 5)(x - 5) (x + 3) (x + 5) (x - 5)(x + 3)

           = (x - 5) (x + 5) (x + 3)


In the page samacheer kalvi math solution for exercise 3.12 part 2  we are going to see the solution of next problem

(iii) 4x² + 9y² + 25z² - 12xy + 30 yz - 20 zx

Solution:

  = (4x² + 9y² + 25z² - 12xy + 30 yz - 20 zx)

  = (2x)²+(-3y)²+(-5z)²+2(2x)(-3y)+2(-3y)(-5z)+2(2x)(-5z) 

  = (2x - 3y - 5z)²

   = (2x - 3y - 5z)


(iv) x⁴ + (1/x⁴) + 2

Solution:

 (a + b)² = a² + 2 a b + b²

             = (x²)² + (1/x²)² + 2 x² (1/x²)]

             = (x² + (1/x²))²

             = (x² + (1/x²))


(v) (6x² + 5 x - 6)(6x² - x - 2)(4x²+ 8 x + 3)

Solution:

            = (6x² + 5 x - 6)(6x² - x - 2)(4x²+ 8 x + 3)

            = √(2x+3)(3x-2)(3x-2)(2x+1)(2x+3)(2x+1)

            = (2x+3)(2x+1)(3x-2)


(vi) (2x² - 5 x + 2)(3x² - 5x - 2)(6x²- x - 1)

Solution:

    = (2x² - 5 x + 2)(3x² - 5x - 2)(6x²- x - 1)

    = (2x-1)(x-2)(3x+1)(x-2)(2x-1)(3x+1)

    = (2x-1)(x-2)(3x+1)

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




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