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Hi,commonsense27 wrote:sqrt(x^2)/x=((x^2)^1/2)/x =x^1-1=x^0=1
It does not matter what is value of x is. Answer is 1. If book is saying otherwise, then book needs to be corrected.
BTW, What is the source of this problem?
sqrt(x^2)/x=((x^2)^1/2)/xcommonsense27 wrote:sqrt(x^2)/x=((x^2)^1/2)/x =x^1-1=x^0=1
It does not matter what is value of x is. Answer is 1. If book is saying otherwise, then book needs to be corrected.
BTW, What is the source of this problem?
The value x^2 is always positive, if x ≠0. And a positive number inside √ is taken as positive only. The value x could be either positive or negative, so the positivity constraint forces us to take √x^2 as |x| only. The answer to this question is -1, if x < 0; and it is 1, if x > 0. Henceforth, two answers are possible, and [spoiler]E[/spoiler] covers both.rahulsaxena8 wrote:
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now what is this in bold?selango wrote:Thanks Sanju.
Rahul,
Sorry I created a lot of confusion here.
The important point to be noted is sqrt(x^2) is always positive whatever the sign of x.
If x=6,sqrt(36)=6/6=1
If x=-6,sqrt(36)/-6=6/-6=-1
Anyways I am glad there are people who actively contribute. I was having tough time understanding why my answer was wrong.selango wrote:Thanks Sanju.
Rahul,
Sorry I created a lot of confusion here.
The important point to be noted is sqrt(x^2) is always positive whatever the sign of x.
If x=6,sqrt(36)/6=6/6=1
If x=-6,sqrt(36)/-6=6/-6=-1
Thank you the thanks' stockpile. I can count 4 from you in this single thread so far. Interesting to see Rahul Lakhani not getting one from you.rahulsaxena8 wrote:Anyways I am glad there are people who actively contribute. I was having tough time understanding why my answer was wrong.selango wrote:Thanks Sanju.
Rahul,
Sorry I created a lot of confusion here.
The important point to be noted is sqrt(x^2) is always positive whatever the sign of x.
If x=6,sqrt(36)/6=6/6=1
If x=-6,sqrt(36)/-6=6/-6=-1
Thanks a lot to everyone who contributed.
No, sqrt(x^2) is *not* always equal to x. You can see this by plugging in x = -2, say. Then sqrt(x^2) = sqrt(4) = 2, which is not equal to x; it's equal to -x. In general, sqrt(x^2) is *always* equal to |x|, but it is only equal to x if x is zero or greater.commonsense27 wrote:I still stand by my answer. If book is saying otherwise, book needs to be corrected.
We should use our fundamentals to arrive at a choice. If everytime we change our fundamentals based on the question, we will be in deep shit.
This is an exponential problem. There is absolutely nothing wrong that I am doing.
Square root cancels the square. And
X^0=1 as long as the x !=0.
ANSWER IS 1.
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