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# a hard question

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the negative reciprocal of the sum of all values of x that satisfy the equation x-sqrt(x+1)/x+sqrt(x+1)=11/5

Below is my steps but idk if its right

36x2 (x to the power of 2) =162 (16 to the power of 2) (x+1)

and then the 16 to the power of to eventually equals 256x+256

Jan 1, 2019

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$$\text{I assume this is (parens are your friend)}\\ \dfrac{x-\sqrt{x+1}}{x+\sqrt{x+1}}=\dfrac{11}{5}$$

$$\text{Let's pull a bit of a trick here}\\ a=\sqrt{x+1}\\ \dfrac{x-a}{x+a}=\dfrac{11}{5}\\ 5x-5a=11x+11a\\ 6x+16a=0\\ 3x=-8a = -8\sqrt{x+1}\\$$

$$9x^2 = 64(x+1)\\ 9x^2 - 64x-64 = 0\\ \text{Using the quadratic formula, (which I leave to you)}\\ x = 8,~-\dfrac{9}{8}\\ \text{Since we squared the equation we must check these roots}$$

$$x=8 \Rightarrow a = 3\\ \dfrac{8-3}{8+3} = \dfrac{5}{11} \neq \dfrac{11}{5}\\ \text{8 is a extraneous solution}\\ x=-\dfrac{8}{9}\Rightarrow a = \dfrac{1}{3}\\ \dfrac{-\dfrac{8}{9}-\dfrac 1 3}{-\dfrac{8}{9}+\dfrac 1 3} = \dfrac{11}{5}\\ \text{so }x=-\dfrac{8}{9} \text{ is an actual root, and the only one}$$

$$-\dfrac{1}{-\dfrac 8 9} = \dfrac 9 8$$

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Jan 1, 2019