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How do you solve #sqrt(2x)=x-4# and find any extraneous solutions?

1 Answer

Konstantinos Michailidis

May 27, 2016

First it should be #x>=0# and #x-4>=0# hence #x>=4#

Now square the relation to get

#2x=(x-4)^2=>2x=x^2-8x+16=> x^2-10x+16=0=>(x-2)*(x-8)=0#

The only acceptable solution is #x=8#

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