-2/x= x/x-2; solve for x?

Answer 1

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

a= 1, b= 2 , c= -4

x = [-2 +/- sqrt(2^2 - 4(1)(-4))] / 2(1)
x = [-2 +/- sqrt(4 + 16)] / 2
x = -1 +/- sqrt(5)

Answer 2

From the eq we get
-2x + 4 = x^2
x^2 + 2x - 4=0
so solving for x,
x= (-2 + 20^(1/2) )/2 & (-2 - 20^(1/2) )/2
= -1 + 5^(1/2) & -1 - 5^(1/2)

Answer 3

-2/x= x/x-2; solve for x?

transpose the fractions

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

ok so lets get the square root of both sides:

sqrt[x^2] = sqrt[-2x+4]

it's obvious that the square root of x to the power of 2 is 2, and the square root of -2x+4 is sqrt[-2x+4]. That's why we conclude therefore that -2/x is not equal to x/x-2 because X is equal to the square root of -2x+4

I hope this helps.. ^_^

Answer 4

- 2(x - 2) = 2 x²
- 2 x + 4 = 2 x²
2x² + 2x - 4 = 0
x² + x - 2 = 0
(x + 2).(x - 1) = 0
x = - 2, x = 1

Answer 5

-2/x=x/(x-2)
=> x*x= -2(x-2) [by cross-multiplication]
=>x^2= -2x+4
=.x^2+2x=4
=>(x)^22*x*1+(1)^2=4+1 [adding 1 to both sides]
=>(x+1)^2=5
=>x+1=+-sqrt 5 [square-rooting both sides]
=>x= -1+-sqrt5

Answer 6

-2(x-2)=x^2
-2x+4=x^2
x^2+2x-4=0
x=-2+sqrt(20)/2 or
x=-2-sqrt(20)/2