Solve 1 - 2/x = 2 / (x+2)?

It is all wrong actually.

in 1 - 2(x + 2) - 2(x) = 0, you have to multiply the whole equation by the denominator of the two fractions, (you left out 1.)

So it should be:

(x+2)(x) - 2x - 2(x + 2) = 0
x^2 + 2x - 2x - 2x - 4 = 0
x^2 - 2x - 4 = 0

Now solve quadratically.

x = (2 +/- ROOT [(-2)^2 - 4 x 1 x -4]) / 2

(2 +/- ROOT [4 + 16]) / 2
(2 +/- ROOT 20) / 2
(2 +/- 2 ROOT 5) / 2 (Because 4 x 5 = 20)

1 +/- ROOT 5

So x is either 3.23, or -1.24 (respectively from 1 + ROOT 5 to 1 - ROOT 5.)

1 - 2/x = 2 / (x+2)
or,x-2=2x/x+2 [multiplying both sides with x ]
or,x^2 -4 = 2x
or, x^2 -2x -4 = 0
or, x^2 -2x +1 -5 = 0
or, (x-1)^2 =5
now, either, x-1=+square root 5
or, x=square root 5 +1.
or, either, x-1 = - square root 5.
or,x= 1- square root 5.

therefore, x=1+square root 5 ,1-square root 5.(ANS)

x = 1 +or - square root of 5
Here's the link in WolframAlpha:

Hi!
1 -(2/x) = 2 /(x+2)
2/(x+2) + 2/x = 1
2x + 2(x+2) = 1(x)(x+2)
2x + 2x + 4 = x^2 + 2x
X^2 -2x - 4 = 0
Solving this quadratic = equation you get the values
x = 1 - sqrt(5) or 1+sqrt(5)

Cheers
Dave

x ( x + 2 ) - 2 ( x + 2 ) = 2 x

x ² + 2 x - 2 x - 4 = 2 x

x ² - 2 x - 4 = 0

x = [ - b ± √ ( b ² - 4 a c ) ] / 2 a

x = [ 2 ± √ ( 4 + 16 ) ] / 2

x = [ 2 ± √ ( 20 ) ] / 2

x = [ 2 ± 2√ 5 ] / 2

x = 1 ± √ 5

1 - 2/x = 2/(x + 2)
x(x + 2)(1 - 2/x) = x(x + 2)[2/(x + 2)]
x(x + 2) - 2(x + 2) = 2x
x^2 + 2x - 2x - 4 = 2x
x^2 - 2x - 4 = 0
x = [-b ±√(b^2 - 4ac)]/(2a)

a = 1
b = -2
c = -4

x = [2 ±√(4 + 16)]/2
x = [2 ±√20]/2
x = [2 ±√(2^2 * 5)]/2
x = [2 ±2√5]/2
x = 1 ±√5

∴ x = 1 ±√5

1 - (2/x)= 2/(x+2)
1- (2/x) - 2/(x+2)=0
multiply the equation by x(x+2) and get
x(x+2) - 2(x+2) - 2x = 0
x^2 + 2x -2x-4 - 2x = 0
x^2 -2x - 4 = 0
then use the quadratic formula

1-2/x=2/(x+2)
(x+2)-2(x+2)/x=2
x^2+4x+4-2x-2=2x
x^2+2=0
x=root2 or -root 2

cut up the bases up: [2^(x) * 2^(a million)] + [2^(x) * 2^(2)] = [2^(-x) * 2^(a million)] + [2^(-x) * 2^(3)] Now potential the phrases which you already know: [2^(x) * 2] + [2^(x) * 4] = [2^(-x) * 2] + [2^(-x) * 8] Multiply the phrases at the same time: 2(2^x) + 4(2^x) = 2(2^-x) + 8(2^-x) whilst the two the bases and exponents are an identical, you could integrate the phrases: 6(2^x) = 10(2^-x) Divide the two aspects via 6: 2^(x) = 10/6(2^-x) Divide the two aspects via (2^-x): [2^(x) / 2^(-x)] = 5/3 whilst dividing phrases with an identical bases, subtract the powers (remember subtracting a adverse is comparable to including): 2^(2x) = 5/3 Take the log of the two aspects: log2(2x) = log(5/3) Divide the two aspects via log2: 2x = [(log5/3) / (log2)] Divide the two aspects via 2: x = [(log5/3) / (2log2)] x = [(log5/3) / (log4)]