How do you solve #\frac { x + 1} { x - 1} + \frac { 2} { x } = \frac { x } { x + 1}#?
1 Answer
Aug 5, 2017
Explanation:
Given:
#(x+1)/(x-1)+2/x=x/(x+1)#
Multiply both sides of the equation by
#x(x+1)^2+2(x-1)(x+1)=x^2(x-1)#
Multiply out to get:
#x^3+2x^2+x+2x^2-2=x^3-x^2#
Add
#5x^2+x-2 = 0#
This is in standard quadratic form:
#ax^2+bx+c = 0#
with
It has solutions given by the quadratic formula:
#x = (-b+-sqrt(b^2-4ac))/(2a)#
#color(white)(x) = (-1+-sqrt(1^2-4(5)(-2)))/(2*5)#
#color(white)(x) = (-1+-sqrt(41))/10#
#color(white)(x) = -1/10+-sqrt(41)/10#
Finally note that these are valid solutions of the original rational equation since neither of them causes any denominator to be zero.