Determine the value of *x* from the following equation:

$\dfrac{1}{x} + \dfrac{1}{x + 2} = 1$

April 19, 2017 - 7:01am

#1
Solve for x: 1/x + 1/(x + 2) = 1

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X= the square root of 2

Simplify the given equation

1/x + 1/(x+2)=1

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

=>(2x+2)/x(x+2)=x(x+2)/x(x+2)

=>2x +2=x^2+2x

=>x^2+2x-2x=2

=>x^2=2

Thus X= the square root of 2 ans.

To verify the answer substitute the value of x=the square root 2 from the given equation if it satisfies equals to 1

Using the calculator by typing the equation using Alpha X and Shift solve it will gives you an answer x= 1.414213562 which is the value of square root of 2.

(1/x) + (1/(x+1)) = 1 multiply by (x * (x+1))

(X+1) + x = x(x+1)

2x + 1 = ×^2 + x

x^2 - x - 1 = 0

x^2 - x = 1

x^2 - x + (1/2)^2 = 1 + (1/2)^2

(x - (1/2))^2 = ( 5/4 )

x - (1/2) = sqrt ( 5/4 )

x = (1/2) +- ((sqrt( 5 ))/2)

X1 = (1 + sqrt ( 5 ))/2

X2 = (1 - sqrt ( 5))/2

+sqrt2 , -sqrt2