SOLUTION: Find the real solutions of the equation. 2x^(-2) - 3x^(-1) - 4 = 0 I then decided to rewrite the original equation this way: (2/x^2) - (3/x) - 4 = 0 Stuck here

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Question 1207661: Find the real solutions of the equation.

2x^(-2) - 3x^(-1) - 4 = 0

I then decided to rewrite the original equation this way:

(2/x^2) - (3/x) - 4 = 0

Stuck here....
Found 4 solutions by Theo, math_tutor2020, ikleyn, Edwin McCravy:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
i did the following:

let k = x^-1

x^-2 = x^-1 * x^-1 = k * k = k^2

2x^(-2) - 3x^(-1) - 4 = 0 becomes:

2k^2 - 3k - 4 = 0

factor by using quadratic formula to get:

k = -.8507810594 or 2.350781059.

since k = x^-1, then:

x^-1 = -.8507810594 or 2.350781059.

this means that 1/k = -.8507810594 or 2.350781059.

solve for k to get:

k = 1/.8507810594 or 1/2.350781059.

k becomes 1.17539053 or .4253905297.

round as necessary.



Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

Multiply both sides by x^2 to go from

to



Rearrange the terms so the quadratic is in standard form.


Now apply the quadratic formula.








or

or

or both of which are approximate.
Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.

Be aware: the solution and the answer by @Theo are incorrect.

One can see it even with unarmed eye.  Indeed,  in  Theo's equation

            2k^2 - 3k - 4 = 0

the leading coefficient  " 2 "  is positive,  while the constant term  " -4 "  is negative.
It implies that the roots of this equation must be with opposite signs;
while Theo gives both roots positive at the end of his solution,  which is wrong.



Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!
   


You can jump right in with the quadratic equation,
solving for x-1, the reciprocal:

 

 

 

Take reciprocals of both sides:

 

Then rationalize the denominator:

Using the +

 



Using the -, similarly 

Edwin
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