krishna kumar wrote:If each expression under the square root is greater than or equal to zero what is root(x^2 - 6x + 9) + root(2 -x) + x -3 ?

All the expression under square root >= 0

When 2 - x >= 0

=> 2 >= x

Thus x is less than equal to 2

When (x^2 - 6x + 9) >= 0

=> (x - 3) ^ 2 >= 0

=> + (x - 3) >= 0 Or - (x - 3) >= 0 (As square root of x = +/- x)

=> x >= 3 Or x <= 3

As we know x is less than equal to 2 we have to consider -(x - 3) as the root of expression (x^2 - 6x + 9) .

root(x^2 - 6x + 9) + root(2 -x) + x -3

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

= root(2 -x)

I myself got to B then I solved this in the manner I wanted to get A. I know its quite impossible to visualize this problem in the way I have shown but I hope it is correct.