How do you solve #\sqrt { 4x - 7} - \sqrt { 5- 2x } = 0#?

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Answer 1 · 2016-10-19T12:02:43

#x=2#

Feasibility implies on

#4x-7 ge 0# and #5-2x ge 0# so

#7/4 le x le 5/2# Now squaring both sidesof

#\sqrt { 4x - 7} = \sqrt { 5- 2x }# we have

#4x-7=5-2x#. Solving for #x# gives

#x = 2# which is a feasible solution for

#7/4 le 2 le 5/2#

Answer 2 · 2016-10-19T12:16:31

#x=2#

#color(blue)(sqrt(4x-7)-sqrt(5-2x)=0#

Add #sqrt(5-2x)# both sides

#rarrsqrt(4x-7)-sqrt(5-2x)+color(red)(sqrt(5-2x))=color(red)(sqrt(5-2x)#

#rarrsqrt(4x-7)cancel(-sqrt(5-2x)+color(red)(sqrt(5-2x)))=color(red)(sqrt(5-2x)#

#rarrsqrt(4x-7)=sqrt(5x-2x)#

Square both sides to remove the radical sign

#rarr(sqrt(4x-7))^(color(red)(2))=(sqrt(5-2x))^(color(red)(2))#

#rarr4x-7=5-2x#

Add #2x# both sides

#rarr4x-7+color(red)(2x)=5-2x+color(red)(2x#

#rarr4x-7+color(red)(2x)=5cancel(-2x+color(red)(2x#

#rarr6x-7=5#

Add #7# to both sides

#rarr6x-7+color(red)(7)=5+color(red)(7)#

#rarr6xcancel(-7+color(red)(7))=5+color(red)(7)#

#rarr6x=12#

Divide both sides by #6#

#rarr(6x)/color(red)(6)=12/color(red)(6)#

#rarr(cancel6x)/cancelcolor(red)(6)=12/color(red)(6)#

#rArrcolor(green)(x=2#