therefore I have actually $sqrta^2+b^2$. I believed that this was same to $a^2+b^2$ but it is not. However, even if I transform the square source to powers, I obtain (based ~ above the power ascendancy $(a^m)^n = a^mn$), I acquire $(a^2+b^2)^0.5 = a^1 + b^1$ but this is still not true...

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

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


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

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

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
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Sambo
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reply Aug 14 "19 in ~ 11:25
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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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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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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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AllawonderAllawonder
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