How many numbers are twice as far from 5 as they are from 8 and what are they?

Question

valpey · Answer

when we say "far" we usually mean something like the absolute value of difference. So let one of these numbers be called "n". Then we'd have:

2|n-5| = |n-8|

valpey · Answer

Now a cool thing about absolute values is that when you square them you can erase the absolute value function and manipulate them with algebra:

$$(2|n-5|)^2 = (|n-8|)^2$$$$2^2(n-5)^2=(n-8)^2$$

valpey · Answer

Now try to solve for n.

anonymous · Answer

i tried that but i got $$3n ^{2}= 39$$ which eventually got me to about 3.6

anonymous · Answer

it just doesnt seem like the right answer

anonymous · Answer

Pepper, you seem to be doing your algebra incorrectly...it should look something like this (starting from Valpey's equation):
$$\begin{align}
&2^2(n-5)^2=(n-8)^2\
&4(n^2-10n+25)=n^2-16n+64\
&4n^2-40n+100=n^2-16n+64\
&3n^2-24n+36=0\
&(3n-18)(n-2)=0\
&n=6,\;2
\end{align}$$

valpey · Answer

Oops..."twice as far from five..." It sounds like it must be:
$$2^2(n-8)^2=(n-5)^2$$instead.

anonymous · Answer

Yep, you're right...this is what I get for not reading the problem.

anonymous · Answer

thanks i just realized that i didnt multiply right stupidd...

anonymous · Answer

wait is n the number of numbers that are twice as far from 5 as they are from 8?

anonymous · Answer

do you guys know if the answer is 7?