How do you solve #x^2/(x^2-4) = x/(x+2)-2/(2-x)#?

1 Answer
Aug 28, 2017

Answer:

#x=2#

Explanation:

Write as:

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

Using the principle that #color(white)("s") a^2-b^2=(a+b)(a-b)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("The sort of cheating bit - not really!")#
Consider #-2/(2-x)#.#color(white)(c)# We need to change this such that the
#color(white)()#
denominator is #x-2# if we whish to use the context of
#color(white)()#
#color(white)("s") a^2-b^2=(a+b)(a-b)#

Suppose we had #+2/(x-2)# which is the format we wish to have. If we use this is it another form of what was originally written. That is; does it have the same value?

#color(green)([2/(x-2)color(red)(xx1)] ->[2/(x-2)color(red)(xx(-1)/(-1))] = -2/((2-x))#

So #+2/(x-2)# has the same value as #-2/(2-x)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Write as:

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

This is the same as:

#x^2/(x^2-2^2)=(x xx2)/((x+2)(x-2))#

#x^2/(x^2-2^2)=(2x)/(x^2-2^2)#

As the denominators are the same we can just consider the numerators.

#x^2=2x#

Divide both sides by x

#x=2#