How do you solve and find the extraneous solutions for #2\sqrt{4-3x}+3=0#?

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Answer 1 · 2014-11-23T15:36:03

#2sqrt{4-3x}+3=0#

by subtracting #3#,

#=> 2sqrt{4-3x}=-3#

by dividing by #2#,

#=> sqrt{4-3x}=-3/2#

(Notice that it is now clear that this equation has no real solution since the left-hand side cannot be negative.)

by squaring,

#=> 4-3x=9/4#

by subtracting #4#,

#=> -3x=-7/4#

by dividing by #-3#,

#=> x=7/12#,

which is its extraneous solution.

I hope that this was helpful.