How to solve `14/(4x^2-1) + 3/(4x^2+4x+1)=5/(4x^2-4x+1)` ?
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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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calendarEducator since 2012
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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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