Hello sinjid9!
Did you make a sketch?
$\displaystyle C$ is a point on line segment $\displaystyle AB$ such that: .$\displaystyle AB\cdot BC \:=\:AC^2$ .[1]
If $\displaystyle AB$ is 6 cm, find the length of $\displaystyle BC$ to the nearest millimeter. Code:
A C B
*-----------*-------*
: 6-x : x :
Let $\displaystyle x \:=\: BC$ . . (Note that: .$\displaystyle 0 \leq x \leq 6$)
Then: $\displaystyle 6-x \:=\:AC$
Substitute into [1]: .$\displaystyle 6x \:=\:(6-x)^2 \quad\Rightarrow\quad x^2 - 18x + 36 \:=\:0$
Quadratic Formula: .$\displaystyle x \;=\; \frac{18 \pm\sqrt{18^2-4\cdot36}}{2(1)} \;=\;\frac{18 \pm6\sqrt{5}}{2} \;=\;9\pm3\sqrt{5}$
Therefore: .$\displaystyle x \;=\; 3(3-\sqrt{5} \;=\;2.291796068\text{ cm} \;\approx\;\boxed{2.3\text{ mm}}$
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A geometric interpretation . . .
Given a line segment $\displaystyle AB$, find an interior point $\displaystyle C$
Code:
A C B
*-----------*-------*
: x : y :
so that these two areas are equal.
Code:
*-------------------*
| |
y | | y
| |
*-----------*-------*
| x | y
| |
x | |
| |
| |
*-----------*
x