How do you solve #sqrt(3x-2)=1+sqrt(2x-3)#?

2 Answers
Sep 29, 2016

Answer:

#x# = 2 and 6

Explanation:

Given:#" "sqrt(3x-2)=1+sqrt(2x-3)#

,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Square both sides

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

The #-2# on both sides cancel out

#3xcancel(-2)" "=" "2xcancel(-2)+2sqrt(2x-3)#

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

Subtract #2x# from both sides

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

Divide both sides by 2

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

Square both sides

#x^2/4 =2x-3#

giving:#" "x^2/4-2x+3=0#

Using the standardised formula #y=ax^2+bx+c#

#a=1/4"; "b=-2"; "c=3#

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

#=> x= (+2+-sqrt((-2)^2-4(1/4)(3)))/(2(1/4))#

#=> x=4 +-(sqrt(4-12/4))/(1/2) #

#x=4+- 2#

#color(brown)(x=2" and "x=6#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("However:")#

Consider #sqrt(2x-3)#

For #x in RR: 2x>=3# otherwise #2x-3# is negative

#color(brown)(=>x>=3/2)#

,..............................................................................................
Consider #sqrt(3x-2)#

For #x in RR: 3x>=2# otherwise #3x-2# is negative

#color(brown)(=> x>=2/3)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Putting it all together")#

#color(brown)("If "x<3/2" then at least one of the roots is negative")#

#color(brown)("As both "x=2" and "x=6" are greater than "3/2)##color(brown)("both are feasible solutions")#

#" "color(green)(bar(ul(|color(white)(2/2)x=2" and "x=6" "|))#
Tony B

Sep 29, 2016

Answer:

#color(green)(x=2color(black" or ")x=6)#

Explanation:

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

Square both sides
#color(white)("XXX")3x-2=1+2sqrt(2x-3)+2x-3#

Simplify
#color(white)("XXX")3x-2=2sqrt(2x-3)+2x-2#

Isolate the term with the root on one side
#color(white)("XXX")x= 2sqrt(2x-3)#

Square both sides
#color(white)("XXX")x^2 = 8x-12#

Convert to standard form
#color(white)("XXX")x^2-8x+12=0#

Factor to get
#color(white)("XXX")(x-2)(x-6)=0#

#rArr x=2 or x=6#