SOLUTION: Find the real roots of the equation 9/x^4+8/x^2=1

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Question 1087964: Find the real roots of the equation 9/x^4+8/x^2=1
Found 3 solutions by math_helper, MathTherapy, ikleyn:
Answer by math_helper(2461) About Me  (Show Source):
Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

USE PARENTHESES if you want a correct answer!

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
9%2Fx%5E4 + 8%2Fx%5E2 = 1 

The standard method to solve it is THIS:


1.  Introduce new variable u = 1%2Fx%5E2.

    Then your equation takes the form

    9u%5E2+%2B+8u+-+1 = 0.

    Factor left side polynomial. You will get

    (9u -1)*(u+1) = 0.

    The roots are  u = 1%2F9  and  u = -1.


2.  Now return to the original  variable x.

    With the root  u = 1%2F9  you have  1%2Fx%5E2 = 1%2F9,

         which implies x = +/- 3.


    With the root  u = -1  you have  1%2Fx%5E2 = -1,

        which HAS NO real solutions.


Answer.  The given equation has two real solutions x = 3 and x = -3.

Solved.