+0

# hard equation

0
85
2

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

Jun 9, 2020

#1
+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
+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