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.

How do you solve and find the extraneous solutions for 2\sqrt{4-3x}+3=02\sqrt{4-3x}+3=0?