How to solve `14/(4x^2-1) + 3/(4x^2+4x+1)=5/(4x^2-4x+1)` ?

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lemjay | High School Teacher | (Level 3) Senior Educator

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`14/(4x^2-1) + 3/(4x^2+4x+1)=5/(4x^2-4x+1)`

First, factor the denominators.

`14/((2x-1)(2x+1))+3/((2x+1)(2x+1)) = 5/((2x-1)(2x-1))`

`14/((2x-1)(2x+1))+3/(2x+1)^2=5/((2x-1)^2)`

Then, multiply both sides by the LCD which is `(2x+1)^2(2x-1)^2` .

`(2x+1)^2(2x-1)^2*[14/((2x-1)(2x+1)) + 3/(2x+1)^2]=5/(2x-1)^2*(2x+1)^2(2x-1)^2`

`14(2x-1)(2x+1)+3(2x-1)^2=5(2x+1)^2`

`14(4x^2-1)+3(4x^2-4x+1)=5(4x^2+4x+1)`

`56x^2-14+12x^2-12x+3=20x^2+20x+5`

`68x^2-12x-11=20x^2+20x+5`

And express the equation in a form `ax^2+bx+c=0` . To do so, subtract both sides by `(20x^2+20x+5)` .

`68x^2-12x-11-(20x^2+20x+5)=20x^2+20x+5-(20x^2+20x+5)`

`48x^2-32x-16=0`

Since 48, 32 and 8 are divisible by 8, divide both sides by 8 to simplify the equation further.

`6x^2-4x-2=0`

Then, factor left side.

`(2x-2)(3x+1)=0`

Set each factor equal to zero and solve for x.

`2x-2=0`                and                      `3x+1=0`

`2x=2`                                                     `3x=-1`

`x=1`                                                         `x=-1/3`

Hece, the solutions to the given equation are `x=1`  and `x=-1/3` .

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tiburtius | High School Teacher | (Level 2) Educator

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`14/(4x^2-1) + 3/(4x^2+4x+1) = 5/(4x^2-4x-1)`

Now let's put everything on the left side and rewrite denominators so we could add the fractions more easily.

`14/((2x-1)(2x+1)) + 3/((2x+1)^2) - 5/((2x-1)^2) = 0`

Now add all 3 fractions ( common denominator is `(2x-1)^2(2x+1)^2`)

`(14(2x-1)(2x+1) + 3(2x-1)^2 - 5(2x+1)^2)/((2x-1)^2(2x+1)^2) = 0`

This fraction will be equal to 0 if numerator is equal to 0. So after we multiply and add everything we get the following quadratic equation

`48x^2 -32x -16 = 0`

Now we devide whole equation with 16 and get

`3x^2 - 2x -1 = 0` 

Now we can solve the equation by using formula for solution of quadratic equation and ultimately we get the following 2 solutions

`x_1 = -1/3` and `x_2 = 1`

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