how to solve this quadratic equation

how to solve this quadratic equation

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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(x^{2} + 4x + 4) = 0

1 + 4x + 8 - 5x^{2} - 20x - 20 = 0

Collecting like terms:

-5x^{2} - 16x - 11 = 0

5x^{2} + 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**

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