what is |CB| in terms of |AB| and |AC| ?
You can solve this get get an expression for |AB| in terms of |AC|.
One term will be negative and thus not a valid length. The other is a valid length.
Divide the valid length by |AC| to get your ratio.
The length of the line segment = x.
The line segment is divided into two parts.
The length of one part = y.
The length of the other part = what in terms of x and y?
You are given that two ratios equal each other. That lets you construct an equation. What is it?
With me so far?
Now solve for x in terms y or y in terms of x. But the answer you are expected to give is a number, not an equation. So what, once you have x in terms of y or y in terms of x, you will take a ratio and BOOM a number will appear.
|AB| = x
|AC| = y
|CB| = x - y
So then it'd be...
$\displaystyle \frac{x}{y} =\frac{y}{x-y}$
Is this what you meant? Would you then do this?:
$\displaystyle (x)(x-y) = y^2$
And if that's correct, I'm not sure what to do next...
Yes. That is correct. You have assigned letters for unknowns and translated the problem into an equation to be solved. That probably represents the first two steps in solving a word problem by algebra 99.9% of the time.
The next step is to solve the equation. Because you probably are most familiar with the quadratic formula expressed in terms of x, I suggest you solve for x in terms of y.
$x(x - y) = y^2 \implies x^2 - xy - y^2 = 0 \implies x = what\ according\ to\ the\ quadratic\ formula.$
Then the last step is to compute the ratio of x to y, using as romsek said, only the relevant solution.
Hello, eleventhhour!
$\displaystyle |AB| = x$
$\displaystyle |AC| = y$
$\displaystyle |CB| = x - y$
Find the ratio $\displaystyle \tfrac{x}{y}$
So then it'd be: .$\displaystyle \frac{x}{y} \,=\,\frac{y}{x-y}$
Is this what you meant? . Yes!
Would you then do this? . $\displaystyle (x)(x-y) \:=\: y^2$ . Yes!
And if that's correct, I'm not sure what to do next.
We have: .$\displaystyle x^2 - xy \:=\:y^2$
. . . . . $\displaystyle x^2 - xy - y^2 \:=\:0$
Divide by $\displaystyle y^2\!:\;\;\frac{x^2}{y^2} - \frac{xy}{y^2} - \frac{y^2}{y^2} \;=\;\frac{0}{y^2}$
. . . . . . . . . $\displaystyle \left(\frac{x}{y}\right)^2 - \frac{x}{y} \;-\;1 \;=\;0$
Let $\displaystyle r = \frac{x}{y}\!:\;\;r^2 - r - 1 \;=\;0$
Quadratic Formula: .$\displaystyle r \;=\;\frac{1 \pm \sqrt{5}}{2}$
Therefore: .$\displaystyle \frac{x}{y} \;=\;\frac{1+\sqrt{5}}{2} \;=\;\phi $ . . . the Golden Mean
Please, please eleventhhour. Pay attention both to the question and to suggestions we make. The site will be more help to you if you do so.
The question DOES NOT ask you to find the length of any segment. It asks you to find the ratio between the lengths of two segments.
I asked you and romsek asked you to find x in terms of y by using the quadratic formula. Did you do that? And then we both told you to calculate x / y. Did you do that? Do you see that x / y is a ratio, which is what you were asked to find? Do you see that when you substitute for x and divide by y, y cancels out and leaves you with a number?
Then soroban came, and in his typically elegant fashion, defined r as x / y, which is a ratio as indicated by the letter he chose to represent it, solved the quadratic in terms of r, and presented you with the only relevant number coming out of that solution.