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December 22, 2014

December 22, 2014

Posted by **Anonymous** on Tuesday, March 19, 2013 at 12:08am.

9x^4+4=13x^2 and

x^6-19x^3=216

Thanks, in advance..

- college algebra -
**saiko**, Tuesday, March 19, 2013 at 2:54amI 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.

- college algebra -
**drwls**, Tuesday, March 19, 2013 at 3:02amThere 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.

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