I wanted to calculate the square root of 80, so I did
$\sqrt{80} = \sqrt {81-1} = 9-1=8$
I do not know what I did wrong, can someone correct me, as $\sqrt{80}$ is about $8.944$.
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8The square root function is not linear. – Xander Henderson Jan 28 '18 at 16:55
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2There are extremely popular threads Pedagogy: How to cure students of the "law of universal linearity?" (on Math.SE) and Whence the "everything is linear" phenomenon, and what can we do about it? (on MathEducators.SE). Out of curiosity, do you have any unique insights into why this phenomenon occurs, having just manifested it? – anon Jan 28 '18 at 17:12
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$$\sqrt{80}=\sqrt{81-1} = 9\sqrt{1-\frac{1}{81}}\approx 9\left(1-\frac{1}{2\cdot 81}\right) = 9-\frac{1}{18}$$ is a correct way to approach it. It is not difficult to realize that $8^2=64\neq 80$, either. – Jack D'Aurizio Jan 28 '18 at 18:16
3 Answers
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It is wrong because $\sqrt{a^2-b^2} \neq a - b$, but $\sqrt{a^2-2ab+b^2} = a-b$ if $a \ge b$.
ArsenBerk
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You cannot just calculate $\sqrt{a^2-b^2}$ as being equal to $a-b$.
This is a common misconception, as if $\sqrt{a^2-b^2}$ was actually $a-b$, this would mean $(a-b)^2=a^2-b^2$, or $a^2-2ab+b^2=a^2-b^2$, which is entirely incorrect (except if $b=0$).
You can only calculate a square root like this if it is in the form of $\sqrt{(a-b)^2}$, which evaluates to $\left(\left(a-b\right)^2\right)^{1/2}$ or $|a-b|$.
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Without digging into algebra here, from a arithmetic standpoint had $\sqrt{80}$ been $8$, then how about $\sqrt{64}$? Clearly those answers can't be the same, can they?
imranfat
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