college algebra
posted by Anonymous .
Solve, what is the easiest way to solve this equation?
9x^4+4=13x^2 and
x^619x^3=216
Thanks, in advance..

I am solving the first one.....second one you can try yourself.
so 9x^4+4=13x^2
this can be written,
x^2*(9x^213)=4
as x^2 is always +ve
so to be ve,
(9x^213)<0
we see that
(sqrt(13)/3)<x<(sqrt(13)/3)
it comes as 1.2<x<1.2
next we will see if there is some integer solution.
at x=1 and x=1
the function has its value 0
so x=1 and x=1 are solution to the equation.
next we will see if there is some maxima or minima in this func,(9x^4+413x^2)
after differentiating,
and taking it =0,
the func. has minima at +sqrt(13/18), sqrt(13/18)
and at x=+1, x=1 it has value 0,
so the another solutions are
x=12*sqrt(13/18) and x=1+2*sqrt(13/18)
these are 4 solutions.
Ask me if you have any doubt. 
There are two equations there.
In the first one, let y = x^2 and solve the quadratic.
9y^2 13y +4 = 0
(y 1)(9y4) = 0
y = 1 or 4/9
x = sqrt(y) = +/1 and +/2/3
In the second one, let y = x^3 and again solve the quadratic.