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
Brent Hanneson - Creator of GMATPrepNow.com
