Question

A point C is chosen on the line segment AB such that AC/BC =(BC)/(5AB). Find (AC)/(BC)

Answer 1

no that was a mistake, let me start again

Answer 2

ok that looks better.
i think it is
\[\frac{1}{\sqrt{5}}\]

Answer 3

i made AB = 1, AC = x and so CB = 1-x then i solved
\[\frac{x}{1-x}=\frac{1-x}{5x}\]
\[5x^2=(1-x)^2\]
\[x=\frac{\sqrt{5}-1}{4}\] then took the ratio
\[\frac{\frac{\sqrt{5}}{4}}{1-\frac{-1+\sqrt{5}}{4}}\]

Answer 4

\[\frac{\sqrt{5}}{4-(-1+\sqrt{5})}\]
\[\frac{-1+ \sqrt{5}}{5-\sqrt{5}}\]
\[\frac{1}{\sqrt{5}}\]

Answer 5

typo there, first line should be
\[\frac{-1+\sqrt{5}}{4-(-1+\sqrt{5})}\]

Answer 6

\[5x ^{2}=(1-x)^{2}\]
\[5x ^{2}=1-2x+x^{2}\]
\[4x ^{2}+2x-1=0\]

By simplifying:
\[x=(-1+\sqrt{2})\div4\]

For AC/BC
\[((-1+\sqrt{2})\div4)\div(1-((-1+\sqrt{2})\div4))\]
\[=(-7+6\sqrt{2})\div3\]

Answer 7

i still think it is
\[5x^2=(1-x)^2\]
\[x=\frac{-1+\sqrt{5}}{4}\] but i may be wrong

Answer 8

no, i am not wrong, it is right
http://www.wolframalpha.com/input/?i=4x^2%2B2x-1%3D0

Answer 9

Yeah you are right. Forgot to multiply the "a" part in the formula b^2-4ac. That is just one silly mistake. Oh boy. :P

Answer 10

There is something I don't like about this problem
AC/BC =(BC)/(5AB)

with Sat's definitions: AB = 1, AC = x and so CB = 1-x
isn't the ratio:
\[ \frac{x}{1-x}= \frac{1-x}{5} \]
Note there is no x in the denominator on the right hand side.
But this ratio leads to negative numbers for x.

Answer 11

@phi Now that you mention it. Your formula is right. There is no 5x involved since 5AB=5.

Answer 12

As an exercise I finished it off this way:
5x= 1-2x+x^2 or x^2-7x+1= 0
this has two roots: (7± 3sqrt(5))/2 of which one leads to positive x:
x= (7-3sqrt(5))/2
and 1-x = (3sqrt(5)-5)/2

the ratio of x/(1-x) = (3sqrt(5)-5)/10 = 0.17082