How do you solve #4/(b+1)= 1 + 2/(2b-1)#?

1 Answer
Oct 11, 2015

Answer:

The given equation has no Real solutions
(see below for Complex solution values)

Explanation:

Given
#color(white)("XXX")4/(b+1)=1+2/(2b-1)#

Multiply both sides by #(b+1)(2b-1)#
#color(white)("XXX")4(2b-1)=(b+1)(2b-1+2)#

Simplify:
#color(white)("XXX")8b-4 = (b+1)(2b+1)#

#color(white)("XXX")8b-4 = 2b^2+3b+1#

Re-write in standard parabolic form
#color(white)("XXX")2b^2-5b+5=0#

Using the quadratic formula:
#color(white)("XXX")b=(5+-sqrt(5^2-4(2)(5)))/(2(2))#

#color(white)("XXX")b= (5+-sqrt(-15))/4#
with no Real roots (because of the negative square root value)
but
Complex roots at #5/4 +-sqrt(15)/4i#

If we re-write
#color(white)("XXX")4/(b+1)=2/(2b-1)+1#
as
#color(white)("XXX")4(b+1)-2/(2b-1)-1 =0#
then examine the graph of left side:
graph{4/(x+1)-2/(2x-1)-1 [-10, 10, -5, 5]}

We see that at no point does it touch the X-axis (i.e. it is never equal to #0#)