How do you solve #1/x + 1 = x/2#?

1 Answer
Feb 27, 2016

Answer:

#x=1+-sqrt(3)#

Explanation:

#1#. In order to solve this equation, each fraction must have the same denominator. To do this, multiply every term by #color(blue)(2x)# since the equation consists of fractions with denominators #color(orange)x# and #color(orange)2#. When you multiply the whole equation by #color(blue)(2x)#, you can get rid of the denominators.

#1/color(orange)x+1=x/color(orange)2#

#color(blue)(2x)(1/x+1)=color(blue)(2x)(x/2)#

#2#. Simplify.

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

#(2color(red)cancelcolor(black)x)/color(red)cancelcolor(black)x+2x=(color(red)cancelcolor(black)2x^2)/color(red)cancelcolor(black)2#

#2+2x=x^2#

#3#. Rewrite the equation in the form, #ax^2+bx+c=0#.

#x^2-2x-2=0#

#4#. Solve for #x# using the quadratic formula.

#color(purple)(a=1),color(turquoise)(b=-2),color(brown)(c=-2)#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-(-2)+-sqrt((-2)^2-4(1)(-2)))/(2(1))#

#x=(2+-sqrt(4+8))/2#

#x=(2+-sqrt(12))/2#

#x=(2+-2sqrt(3))/2#

#x=(2(1+-sqrt(3)))/(2(1))#

#x=(color(red)cancelcolor(black)2(1+-sqrt(3)))/(color(red)cancelcolor(black)2(1))#

#color(green)(x=1+-sqrt(3))#

#:.#, #x=1+-sqrt(3)#.