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Solve \(\dfrac{1}{2x - 1} + \dfrac{1}{2x + 1} + \dfrac{7}{4x^2 - 1} = 1\)

 Jun 9, 2020
 #1
avatar+26620 
+2

Note that 4x^2 -1  =  (2x-1)(2x+1)      multiply BOTH sides of the equation by this to get

 

2x+1     + 2x-1   + 7   = 4x^2 -1     simplify

4x + 7 = 4x^2 -1

4x^2-4x-8 = 0            Quadratic Formula yields   x = 2  or  -1

 Jun 9, 2020
edited by ElectricPavlov  Jun 9, 2020
edited by ElectricPavlov  Jun 9, 2020
 #2
avatar+131 
+5

Your not wrong when you say this is hard...

 

I'll take a shot at it...

 

I got the values x = 2

and x = -1

 

They seem to work...

 

I first factored 4x^2 - 1 to get (2x + 1)(2x - 1). When we add fractions we multiply the denominators to get the common denominator.

 

When I factored 4x^2 - 1 I got (2x + 1)(2x - 1) and that is the same denominator when you multiply the other 2 denominators together.

 

We get 4x/(2x + 1)(2x - 1)

 

We multiply 1 by (2x + 1)(2x - 1) so the denominator cancels out.

 

We get 4x + 7 = (2x + 1)(2x - 1)

 

We expand (2x + 1)(2x - 1) to get 4x^2 - 1. 

 

4x + 7  = 4x^2 - 1

 

We subract 7 on both sides to receive 4x^2 - 8 = 4x

 

Then, we subtract 4x from both sides to get 4x^2 - 8 - 4x = 0

 

We can use the quadratic formula to find the solution from here on out.

 

a = 4, b = -4, c = -8

 

\(x = {-4 + \sqrt{(-4)^2-4(4(-8)} \over 2(4)}\)

 

Saving you time, this gives us 2 and -1.

 Jun 9, 2020

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