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

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Answer 1 · 2014-11-23T15:36:03

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

by subtracting $3$ $3$ ,

$\Rightarrow 2 \sqrt{4 - 3 x} = - 3$ $=> 2sqrt{4-3x}=-3$

by dividing by $2$ $2$ ,

$\Rightarrow \sqrt{4 - 3 x} = - \frac{3}{2}$ $=> sqrt{4-3x}=-3/2$

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

by squaring,

$\Rightarrow 4 - 3 x = \frac{9}{4}$ $=> 4-3x=9/4$

by subtracting $4$ $4$ ,

$\Rightarrow - 3 x = - \frac{7}{4}$ $=> -3x=-7/4$

by dividing by $- 3$ $-3$ ,

$\Rightarrow x = \frac{7}{12}$ $=> x=7/12$ ,

which is its extraneous solution.

I hope that this was helpful.

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