Posted by Anonymous on Tuesday, March 19, 2013 at 12:08am.
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^2-13)=-4
as x^2 is always +ve
so to be -ve,
(9x^2-13)<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+4-13x^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=1-2*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)(9y-4) = 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.