How do you solve #sqrt(2X-1) = 5 - sqrt(X-1)# and find any extraneous solutions?

1 Answer
Mar 14, 2017

Answer:

#color(green)(x=5)#
#color(white)("XXX")#extraneous solution at #x=145#

Explanation:

Given
#color(white)("XXX")sqrt(2x-1)+sqrt(x-1)=5#

Squaring both sides
#color(white)("XXX")2x-1+2sqrt(2x-1)sqrt(x-1)+x-1=25#

Simplifying
#color(white)("XXX")3x-2+2sqrt(2x^2-3x+1)=25#

#color(white)("XXX")2sqrt(2x^2-3x+1)=27-3x#

Squaring both sides again
#color(white)("XXX")8x^2-12x+4=729-162x+x^2#

Simplifying into standard quadratic form
#color(white)("XXX")x^2-150x+725#

Factoring (or use the quadratic formula if the factors are not apparent)
#color(white)("XXX")(x-5)(x-145)=0#

Giving possible solutions:
#color(white)("XXX")x=5color(white)("XX")andcolor(white)("XX")x=145#

Substituting back into the original equation:
[1]#color(white)("XXX")x=5#
#color(white)("XXXXX")LS#
#color(white)("XXXXXXX")=sqrt(2*5-1)#
#color(white)("XXXXXXX")=sqrt(9)#
#color(white)("XXXXXXX")=3#
#color(white)("XXXXX")RS#
#color(white)("XXXXXXX")=5-sqrt(5-1)#
#color(white)("XXXXXXX")=5-2#
#color(white)("XXXXXXX")=3#
#color(white)("XXX")LS=RScolor(white)("XXX")#valid

[2]#color(white)("XXX")x=145#
#color(white)("XXX")LS#
#color(white)("XXXXX")=sqrt(2*145-1)#
#color(white)("XXXXXXX")=sqrt(289)#
#color(white)("XXXXXXX")=17#
#color(white)("XXX")RS#
#color(white)("XXXXXXX")=sqrt(145-1)#
#color(white)("XXXXXXX")=sqrt(144)#
#color(white)("XXXXXXX")=12#
#color(white)("XXX")LS!=RScolor(white)("XXX")#extraneous