What is the value of √x^2
Answers
The answer of the question is:
X
Answer:
Why does x2−−√=|x| (absolute value) while (x−−√)2=x ?
So why does x2−−√=|x| but ( x−−√)2=x ?
In short, it’s because the square root function always selects the positive root. That’s why there’s the absolute value in the first equation. The absolute value goes away in the second equation because it is squared, which eliminates the sign distinction (e.g. (−c)2=(−1)2c2=c2 ).
Before jumping into the detailed explanation observe that if x≥0 it doesn’t really matter whether we use |x| or x.
32−−√=9–√=3
3–√2=3
But if x < 0 we see the difference.
(−3)2−−−−−√=9–√=3=|−3|
−3−−−√2=(3–√i)2=−3
Why does it work if x≥0 but not if x<0 ? Because the square root function selects (somewhat arbitrarily) the positive root, not the negative one. Both (−3)2=9 and 32=9 . However 9–√=3 , not -3. To get the negative root you need to negate the positive one. E.g. −9–√=−3 gives the second root.
What does it mean that the square root function arbitrarily selects the positive root? Well you could just as easily define a function, let’s call it N, that selects the negative root. Then we’d have
N(32)=N(9)=−3
N(3)2=3
And
N((−3)2)=N(9)=−3
N(−3)2=(N(3)i)2=−3
So when we use a function that selects the negative root it works out fine in the negative case but not in the positive case. In this sense the choice of N isn’t any better or worse than the choice of √ . However √ does have some notational conveniences when it comes to positive numbers.
81−−√−−−−√=9–√=3
N(N(81))=N(−9)=−3i
−N(−N(81))=−N(9)=−(−3)=3
Simply chaining N’s will result in complex values due to the negative square root. It requires the additional specification of the positive root (-N(x)) to avoid this.
Rather than specify a particular root it’s possible to specify both roots simultaneously using the ± symbol. Suppose x2=9 . Then x=±3 gives both solutions simultaneously. Note that some care needs to be taken with this notation. Specifically one may need to take care when negating the equations (sometimes written as ∓ ) or incorporating multiple ± signs.
The other answers saying that this is because the square root is defined as the positive square root are proximately correct. Since f(x)=x2 is an even function then f(−x)−−−−−√=f(x)−−−−√ so f(x)−−−−√=f(|x|)−−−−−√ , and thus since the input to f was in the restricted domain of the partial inverse, we get x2−−√=|x| . But there is a deeper and more general reason why that no matter how we define x−−√ , x2−−√ cannot equal x !
The deeper reason is that the squaring function is not injective (not “one-to-one”, i.e. you can find two numbers which square to the same number), which is why we have to make a choice in the first place. In general, if f is a non-injective function and f−1 is a “partial inverse”, that is, a function which is an inverse of f restricted to a domain on which it is injective, then f−1∘f≠id in general, where id is the identity function. The reason for this is that if x and y are two inputs for which x≠y but f(x)=f(y) , then if we denote this common value by z , so z=f(x)=f(y) , we have that (f−1∘f)(x)=f−1(f(x))=f−1(z) but (f−1∘f)(y)=f−1(f(y))=f−1(z) as well. Yet this means then (f−1∘f)(x)=(f−1∘f)(y) , which means that at least one of them has to be something other than x or y , since those are not equal! Thus f−1∘f cannot be the identity function, and so if f(x)=x2 and f−1(x)=x−−√ , then it cannot be true that x2−−√=x … no matter what choice we make for the square root! And for reasons already given, it turns out that more specifically, x2−−√=|x| .
The principal square root is the positive square root of a number. The absolute value function makes every number positive, regardless what number is inside.
|5|=+5
|−3|=+3
|0|=+0
|−(−(−|−4|)|=+4
For positive numbers (that is strictly positive or zero), there’s no difference because |x|=x when x is positive.
32−−√=9–√=3
3–√2=3
692−−−√=4761−−−−√=69
69−−√2=69
For strictly negative numbers, however, it’s obvious that x2=|x|≠x :
(−1)2−−−−−√=1–√=1≠−1
−1−−−√2=? (square roots of negative numbers are not even defined on the real number line)
We should extended the field to the complex plane:
(−1)2−−−−−√=1–√=1≠−1
−1−−−√2=(±i)2=−1
The function y=x2 doesn’t have an inverse. If x2=y , x=±y√ which doesn’t produce a unique value.