Question

Plz any one give me the answer its urgent ​

Answer 1

1. Solution :

=> x² - 4√2x + 6 = 0

=> x² - √2x - 3√2x + 6 = 0

=> x(x - √2) - 3√2(x - √2) = 0

=> (x - √2)(x - 3√2) = 0

=> x = √2 , 3√2

2. Solution :

=> 3x² + 11x + 10 = 0

=> 3x² + 6x + 5x + 10 = 0

=> 3x(x + 2) + 5(x + 2) = 0

=> (x + 2)(3x + 5) = 0

=> x = -2 , -5/3

3. Solution :

=> => 2x² + x + 4 = 0

=> 2(x² + x/2 + 2) = 0

=> x² + x/2 + 2 = 0

=> x² + x/2 + (1/4)² - (1/4)² + 2 = 0

=> x² + 2•x•(1/4) + (1/4)² = (1/4)² - 2

=> (x + 1/4)² = 1/16 - 2

=> (x + 1/4)² = (1 - 32)/2

=> (x + 1/4)² = -31/2

This is not possible that square of any real numbers is negative , thus there is no real value of x which satisfy the given equation .