# Intersection points

by SeReNiTy
Tags: intersection, points
 P: 171 Hi guys, i'm just wondering is it possible to solve the following using algebra to obtain the points of intersection of the two curves f(x) = 6sqrt(x) and g(x) = [(x+5)^2]/36 I got to the point where i reconized that the inverse of g(x) = 6sqrt(x) - 5 which looks alot like the function f(x), any hints or solutions to this problem?
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,502 Yes, the inverse of g is f(x)- 5. I don't know that that helps a lot in finding points of intersection. The only way I see of finding the points of intersection is to set f(x)= g(x), square both sides to get rid of the square root, and solve the resulting fourth degree equation
 HW Helper P: 1,021 I don't know if you can get there your way (something with the inverse) but tehcnically, it's solvable since you get a 4th-degree polynomial. It won't be 'fun' though. Of course, watch out for introducing solution when squaring. $$\frac{{\left( {x + 5} \right)^2 }} {{36}} = 6\sqrt x \Leftrightarrow \left( {\frac{{\left( {x + 5} \right)^2 }} {{36}}} \right)^2 = \left( {6\sqrt x } \right)^2 \Leftrightarrow \frac{{\left( {x + 5} \right)^4 }} {{1296}} - 36x = 0$$ If you'd want to know, mathematica gives me: $$\begin{gathered}-5 + \frac{{\sqrt{\frac{{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}}}{3} + 72\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}}}}}{2} - \hfill \\ \frac{{\sqrt{\frac{-{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}} - 216\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}} + \frac{279936} {{\sqrt{\frac{{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}}}{3} + 72\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}}}}}} {3}}}}{2}\end{gathered}$$ and $$\begin{gathered}-5 + \frac{{\sqrt{\frac{{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}}}{3} + 72\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}}}}}{2} + \hfill \\ \frac{{\sqrt{\frac{-{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}} - 216\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}} + \frac{279936} {{\sqrt{\frac{{\left( 29386561536 - 120932352\,{\sqrt{57549}} \right) }^{\frac{1}{3}}}{3} + 72\,2^{\frac{2}{3}}\,{\left( 3\,\left( 243 + {\sqrt{57549}} \right) \right) }^{\frac{1}{3}}}}}} {3}}}}{2}\end{gathered}$$ which are approx: $x \to 0.013452\, \wedge \,x \to 29.150$
P: 171
Intersection points

 Quote by HallsofIvy Yes, the inverse of g is f(x)- 5. I don't know that that helps a lot in finding points of intersection. The only way I see of finding the points of intersection is to set f(x)= g(x), square both sides to get rid of the square root, and solve the resulting fourth degree equation
Oh, with the inverse thing, i just remembered doing a question a while ago that involved finding the points of intersection between two inverse functions, it was much easier to computer the intersection between one of the functions and y = x since inverse always intersect along that line.

How do i go about solving the 4th degree polynomial?
HW Helper
P: 1,021
 Quote by SeReNiTy How do i go about solving the 4th degree polynomial?
Just as there are formula's for the quadratic and cubic, there also exists one for the 4th-degree, named Ferrari.

http://mathworld.wolfram.com/QuarticEquation.html

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