You are watching: Sqrt(a^2-b^2)

Why is this, and is there any kind of other dominion for simple $sqrta^2+b^2$?

It is true that (with part restriction):$$ (a^m)^n = a^mn$$

It is likewise true that:$$ <(ab)^m>^n = ^n=

You say:

$(a^2+b^2)^0.5 = a^1 + b^1$

However **This is not a general rule** once you have actually "addition" operation raised to a power. In this details case that is true at least when $a=b=0$

In situation you have $(x+y)^m$, where $m$ is an confident integer, there is an development for this making use of the Binomial Theorem.

In instance you have actually $(x+y)^r$, whereby $r$ is not an integer, over there is one infinite series for this instance using numerous approximation techniques such as Taylor Expansion. There is additionally a binomial development for fractional Exponents.

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edited Aug 14 "19 in ~ 12:17

Sambo

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reply Aug 14 "19 in ~ 11:25

NoChanceNoChance

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In power rule that you mentioned, specific $(a^m)^n=a^mn$, $a^m$ is a single number, whereas in $(a^2+b^2)^0.5$, the 0.5"th strength is used to a sum, for this reason this is a different case.

To watch that $sqrta^2+b^2=a+b$ is really false, find a counterexample. Take a=3 and also b=4 for example.

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answered Aug 14 "19 in ~ 10:48

ScientificaScientifica

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Your an initial attempt quantities to

$$sqrt s=s$$ which is clearly wrong.

Your second attempt does not fit with the strength rule.

$$sqrts^2=s^1/2cdot2=s$$ would certainly be right, however is no what girlfriend considered.

Now have a look at

$$sqrt1+t^2$$ and shot to somehow relate it come $t$.

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reply Aug 14 "19 at 10:44

user65203user65203

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It is very tempting to assume the $(a+b)^2$ is same to $a^2 + b^2$, when, in fact, it is not.

Thus $sqrta^2+b^2$ is not equal to $a+b$. $sqrta^2 + b^2$ is around as simplified as you deserve to go.

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answered Aug 14 "19 at 17:57

MarvinMarvin

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I would like to start with a principle that need to be taught in every schools:

**In aramuseum.orgematics, naught Is True** unless there’s a proof the it’s true.

There space a the majority of formulas that look really pretty and seem really reasonable, and are true besides, favor $(ab)^n=a^nb^n$, but you should have actually been shown in college *why* that formula is true.

You were undoubtedly hoping that the same pretty and reasonable formula $(a+b)^n=a^n+b^n$ would certainly be true, yet there’s no proof because that this. In fact, it’s **false**, however we have actually something much better, a formula through a stern and also crystalline beauty beauty of the own, referred to as the Binomial Theorem:$$(a+b)^n=a^n+na^n-1b + fracn(n-1)2a^n-2b^2+cdots+fracn!(n-j)!j!a^n-jb^j+cdots+nab^n-1+b^n,,$$valid once $n$ is a positive entirety number.

The moral of my sermon? nothing ask why an equation or formula no true, due to the fact that most space not true. Rather, ask what *is* true.

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reply Aug 14 "19 at 20:45

LubinLubin

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Answering the titular questions, i m sorry are very clear:

The reason that $sqrta^2+b^2 e a +b$ is that when you square $a+b$ friend don"t get ago $a^2+b^2$ together you should have if undoubtedly that to be the square root. In fact, you have actually instead $(a+b)^2=a^2+2ab+ b^2,$ which is turn off by the ax $2ab.$

Well, there"s no other way given every we recognize that we can simplify $sqrta^2+b^2$ further. It"s the square root of a amount of two squares, and if $a$ and $b$ are confident it to represent the size of the hypotenuse the a best triangle with legs that lengths $a$ and also $b.$

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reply Aug 14 "19 in ~ 18:10

AllawonderAllawonder

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