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

Nov 23, 2014

$2 \sqrt{4 - 3 x} + 3 = 0$

by subtracting $3$,

$\implies 2 \sqrt{4 - 3 x} = - 3$

by dividing by $2$,

$\implies \sqrt{4 - 3 x} = - \frac{3}{2}$

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

by squaring,

$\implies 4 - 3 x = \frac{9}{4}$

by subtracting $4$,

$\implies - 3 x = - \frac{7}{4}$

by dividing by $- 3$,

$\implies x = \frac{7}{12}$,

which is its extraneous solution.

I hope that this was helpful.