
#1
Aug705, 05:35 AM

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? 



#2
Aug705, 05:44 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,877

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 



#3
Aug705, 05:45 AM

HW Helper
P: 1,024

I don't know if you can get there your way (something with the inverse) but tehcnically, it's solvable since you get a 4thdegree polynomial. It won't be 'fun' though. Of course, watch out for introducing solution when squaring.
[tex]\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[/tex] If you'd want to know, mathematica gives me: [tex]\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} [/tex] and [tex]\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} [/tex] which are approx: [itex]x \to 0.013452\, \wedge \,x \to 29.150[/itex] 



#4
Aug705, 05:48 AM

P: 171

Intersection pointsHow do i go about solving the 4th degree polynomial? 



#5
Aug705, 05:54 AM

HW Helper
P: 1,024

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


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