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

(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)|

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