Can you please show work? Thanks! *Note the x's are not part of the radicand Solve for x

Question

#sqrt2 x(3sqrt2 x-6)=0#

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Answer 1 · 2018-03-01T00:00:58

#x=0, x=sqrt(2)#

Apply the Distributive Property, #a(b+-c)=ab+-ac#

#sqrt(2)x(3sqrt(2)x-6)=sqrt(2)x*3sqrt(2)x-sqrt(2)x*6#

#3sqrt(2)^2x^2-6sqrt(2)x=0#

#3(2)x^2-6sqrt(2)x=0# (Because #sqrt(a)^2=a#)

#6x^2-6sqrt(2)x=0#

Both of our terms include an instance of #6x#, so we can factor out #6x:#

#6x(x-sqrt(2))=0#

We now have two equations two solve for #x:#

#6x=0# and #x-sqrt(2)=0#

#6x=0#

#(cancel6x)/cancel6=0/6#

#x=0#

So, #x=0# is one of our answers.

#x-sqrt(2)=0#

#xcancel(-sqrt(2)+sqrt(2))=0+sqrt(2)#

#x=sqrt(2)#

So, #x=sqrt(2)# is our other answer.