sana.noor wrote:If each expression under the square root is greater than or equal to 0, what is √(x^2 - 6x + 9) + √(2 - x) + x - 3?

a. √(2-x)
b. 2x - 6 + √(2-x)
c. √(2-x) + x - 3
d. 2x - 6 + √(x-2)
e. x + √(x-2)

Here's the algebraic approach.

√(x² - 6x + 9) + √(2 - x) + x - 3 = √[(x-3)(x-3)] + √(2 - x) + x - 3
= √(x-3)² + √(2 - x) + x - 3

IMPORTANT THING #1: all expressions under square root sign are greater than or equal to 0.
Notice that we have the expression √(2 - x)
If 2 - x > 0, then we know that x < 2

IMPORTANT THING #2:
Notice that √k² = k or -k, depending on the value of k.
If k = 5, then √k² = 5 (= k)
If k = -5, then √k² = 5 (= -k)
So, √k² = k or -k since the square root of a value must be greater than or equal to zero.

Similarly, √(x-3)² = (x-3) or -(x-3)
In other words, √(x-3)² = (x-3) or -x+3, depending on whether (x-3) is positive or negative.

Since we already determined that x < 2, we can conclude that (x-3) is negative and (-x + 3) is positive.
Since the square root of a value must be greater than or equal to zero it must be the case that √(x-3)² = -x + 3

Now back to our simplification...
√(x-3)² + √(2 - x) + x - 3
= -x + 3 + √(2 - x) + x - 3
= √(2 - x)
= A

Cheers,
Brent