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

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Answer 1 · 2018-04-08T09:38:33

#x=8#

#sqrt(x+1)+1=sqrt(2x), x>0#

#1=sqrt(x+1)-sqrt(2x)# subtract #sqrt(x-1)# from both sides

#1=2x-2sqrt(2x(x+1))+(x+1)# square both sides

#0=3x-2sqrt(2x(x+1))# collect terms and subtract #1# from both sides

#sqrt(2x(x+1))=(3x)/2# add #sqrt(2x(x+1))# to both sides

#2x(x+1)=(9x^2)/4# square both sides

#2x^2+2x=(9x^2)/4# expand brackets

#0=x^2/4-2x# move terms to RHS

#0=x(x/4-2)# factorise

#x=0 or x=8# solve linear equations

Since #x>0#, #x=8#

Answer 2 · 2018-04-08T10:00:43

#x=8#

Given:

#sqrt(x+1)+1 = sqrt(2x)#

Square both sides (noting that squaring can introduce extraneous solutions) to get:

#(x+1)+2sqrt(x+1)+1 = 2x#

Subtract #x+2# from both sides to get:

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

Square both sides to get:

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

Subtract #4x+4# from both sides to get:

#0 = x^2-8x = x(x-8)#

So #x=0# or #x=8#

Check whether these are solutions of the original equation:

#sqrt((color(blue)(0))+1)+1 = 2 != 0 = sqrt(2(color(blue)(0)))#

#sqrt((color(blue)(8))+1)+1 = 3+1 = 4 = sqrt(16) = sqrt(2(color(blue)(8)))#

So #x=8# is the only solution of the given equation.