Is SQRT( (x-5)^2 =5-x?
1) -x|x|>0
2)5-x>0
Ans- D ( BOTH SUFFICIENT)
Source: Gmatprep 2
Please explain?
mbarshaik Wrote:SQRT( (x-5)^2 =5-x?
SQRT ( (x-5)^2 = |x-5|
|x-5| = 5-x
Now the above equation has two values
x-5 = 5-x or -(x-5) = 5-x depending on whether (x-5) > or < 0
1. -x|x| > 0
the above is possible only for x < 0, therefore (x-5) < 0
2. Clearly states that (x-5) < 0 so this is sufficient and the answer is D
veronica.tong Wrote:Ron,
Can you please explain what you mean by:
if it's to be the original quantity (x - 5), then that quantity must be at least 0: x > 5.
if it's to be the opposite (5 - x), then that opposite quantity must be at least 0. for that to happen, x < 5.
(notice that, if x is actually 5, then |x - 5| equals both (x - 5) and (5 - x), since both of them are zero.)
Why must the left side of the equation be at least zero? Where did you get that information from?
Been reading your explanation over and over again, I get everything else except for that part. I understand that if you plug in numbers, like -6 vs. 6, the answer is very obvious, but I want to know how you can solve this problem without doing that. Please help! Thank you!
RonPurewal Wrote:general takeaway here:
squaring a quantity, and then square-rooting, is equivalent to taking the absolute value.
remember this.
RonPurewal Wrote:|x-5| = 5-x
Now the above equation has two values
x-5 = 5-x or -(x-5) = 5-x depending on whether (x-5) > or < 0
we can make this more clear:
|x - 5| can be either (x - 5), the actual quantity within the absolute-value bars, or (5 - x), the opposite of that quantity.
if it's to be the original quantity (x - 5), then that quantity must be at least 0: x > 5.
if it's to be the opposite (5 - x), then that opposite quantity must be at least 0. for that to happen, x < 5.
(notice that, if x is actually 5, then |x - 5| equals both (x - 5) and (5 - x), since both of them are zero.)
therefore, we can rephrase the question:
is x < 5?
RonPurewal Wrote:we can make this more clear:
|x - 5| can be either (x - 5), the actual quantity within the absolute-value bars, or (5 - x), the opposite of that quantity.
if it's to be the original quantity (x - 5), then that quantity must be at least 0: x > 5.
if it's to be the opposite (5 - x), then that opposite quantity must be at least 0. for that to happen, x < 5.
(notice that, if x is actually 5, then |x - 5| equals both (x - 5) and (5 - x), since both of them are zero.)
RonPurewal Wrote:therefore, we can rephrase the question:
is x < 5?
supratim7 Wrote:1) -x|x| > 0
so, x < 0
any negative number satisfies |x-5| = 5-x
2) 5-x > 0
so, x < 5
any number less than 5 satisfies |x-5| = 5-x
(C) If the question is "Is √{(x-5)^2} = 5-x ?", Why cant we square both side and solve? i.e.
Is √{(x-5)^2} = 5-x ?
Is (x-5)^2 = (5-x)^2 ?