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how to solve (1/(x+2))^2+(4/(x+2))-5=0

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3 Answers

Let u = 1/(x+2)

u^2 + 4u - 5 = (u+5)(u-1) = 0

u = -5 = 1/(x+2) => x+2 = -1/5 => x = -11/5

or

u = 1 = 1/(x+2) => x+2 = 1 => x = -1

Answer: x = -1, -11/5 {edited}

Since you have variables on the bottom of fractions, you need to multiply through the entire equation by the least common denominator (LCD). In this case, your LCD is (x+2)2.

[(x+2)2] {[1/(x+2)]2+[4/(x+2)]-5=0}

1 + 4(x+2) - 5(x+2)2 = 0

1 + 4x + 8 - 5(x2 + 4x + 4) = 0

1 + 4x + 8 - 5x2 - 20x - 20 = 0

Collecting like terms:

-5x2 - 16x - 11 = 0

5x2 + 16x + 11 = 0

(5x + 11)(x + 1) = 0

5x+11 = 0 OR x+1 = 0

x = -11/5 OR x = -1

     1/(x + 2)2  +  4/(x + 2)  -  5  =  0

     (x + 2)-2  +  4(x + 2)-1  -  5  =  0

     ((x + 2)-1  +  5)ยท((x + 2)-1  -  1)  =  0

          (x + 2)-1  +  5  =  0          and          (x + 2)-1  -  1  =  0

     (x + 2)-1  +  5  =  0

     1/(x + 2)  +  5  =  0

     1/(x + 2)  =  -5

     -5(x + 2)  =  1

     -5x  -  10  =  1

     -5x  =  11

      x  =  11/-5    

     x  =  -11/5

AND

     (x + 2)-1  -  1  =  0

     1/(x + 2)  -  1  =  0

     1/(x + 2)  =  1

     x + 2  =  1

     x = -1