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

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons -> 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      Log On


   



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) About Me  (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) About Me  (Show Source):
You can put this solution on YOUR website!

Multiply both sides by x^2 to go from
2%2F%28x%5E2%29+-+3%2Fx+-+4+=+0
to
2+-+3x+-+4x%5E2+=+0


Rearrange the terms so the quadratic is in standard form.
-4x%5E2+-+3x+%2B+2+=+0

Now apply the quadratic formula.
x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x+=+%28-%28-3%29%2B-sqrt%28%28-3%29%5E2-4%28-4%29%282%29%29%29%2F%282%28-4%29%29

x+=+%283%2B-sqrt%289+%2B+32%29%29%2F%28-8%29

x+=+%283%2B-sqrt%2841%29%29%2F%28-8%29

x+=+%283%2Bsqrt%2841%29%29%2F%28-8%29 or x+=+%283-sqrt%2841%29%29%2F%28-8%29

x+=+%28-3-sqrt%2841%29%29%2F%288%29 or x+=+%28-3%2Bsqrt%2841%29%29%2F%288%29

x+=+-1.175391 or x+=+0.425391 both of which are approximate.

Answer by ikleyn(52776) About Me  (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) About Me  (Show Source):
You can put this solution on YOUR website!
   
2x%5E%28-2%29+-+3x%5E%28-1%29+-+4+%22%22=%22%22+0

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

 x%5E%28-1%29+%22%22=%22%22+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

 x%5E%28-1%29+%22%22=%22%22+%28-%28-3%29+%2B-+sqrt%28%28-3%29%5E2-4%2A2%2A%28-4%29+%29%29%2F%282%2A%28-2%29%29+

 x%5E%28-1%29+%22%22=%22%22+%283+%2B-+sqrt%2841%29%29%2F4+

Take reciprocals of both sides:

 x+%22%22=%22%22+4%2F%283+%2B-+sqrt%2841%29%29

Then rationalize the denominator:

Using the +

x+%22%22=%22%22 

x+%22%22=%22%22+4%283-sqrt%2841%29%29%2F%289-41%29%22%22=%22%224%283-sqrt%2841%29%29%2F%28-32%29%22%22=%22%22%28-%283-sqrt%2841%29%29%29%2F8%22%22=%22%22%28-3%2Bsqrt%2841%29%29%2F8

Using the -, similarly x+%22%22=%22%22%28-3-sqrt%2841%29%29%2F8

Edwin