Why isn't the $\sqrt{80}=8$?

Question

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 $8.944$.

Answer 1

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 $|a-b|$.

Answer 2

It is wrong because $\sqrt{a^2-b^2} \neq a - b$, but $\sqrt{a^2-2ab+b^2} = a-b$ if $a \ge b$.

Answer 3

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?