We have $f(x)=x^2-3x-2$ where $x\ge3$. graph of two functions $y=f^{-1}(x)$ and $g(x)=x-3$ intersect each other at the point $A$. What is the distance from the point $A$ to the origin?
$1)\sqrt{74}\qquad\qquad\qquad2)\sqrt{69}\qquad\qquad\qquad3)\sqrt{89}\qquad\qquad\qquad4)\sqrt{97}$
It is a problem from a timed exam, so I am looking for the fastest answers.
One approach is completing the square of $x^2-3x-2$ then find $f^{-1}(x)$ and equate it with $g(x)=x-3$, but since completing the square containing some fractions it seems not a good idea to solve the problem quickly.
Another approach: I noticed that $f(x)$ and $y=x$ intersect each other at the same point(s) as $f^{-1}(x)$ intersects $y=x$ (because $f$ and $f^{-1}$ are symmetrical along $y=x$) so we can find the points of intersections of $f^{-1}(x)$ and $y=x$ by solving the equation $x^2-3x-2=x$, but I can't make a connection between those points and points of intersections of $f^{-1}(x)$ and $y=x\color{red}{-3}$.
The inverse function is given by $y$ where $x=y^2-3y-2$ (just swap the x and the y).
Line is $y=x-3$
Write that as $x=y+3$ and equate the two;
$y^2-3y-2=y+3$
Rearrange and solve easy quadratic.