Question

Which of the following is an extraneous solution of sqrt(-3x-2)=x+2 a.-6 b.-1 c.1 d.6

Answer 1

Try plugging in the 4 solutions and see if they fit
for example a, x - 6 is not extraneous  because  

left side = sqrt (-3*-6 - 2) == sqrt16 = 4 or -4
right side = -6 + 2 = -4

so x = -6 is not extraneous

Answer 2

Answer:

- 6 is the extraneous solution.

Step-by-step explanation:

Given : $\sqrt{-3x -2} = x + 2$.

To find : Which of the following is an extraneous solution .

Solution : We have given that $\sqrt{-3x -2} = x + 2$.

Taking square both sides

-3x - 2 = $(x+2)^{2}$.

On applying identity  $(a+b)^{2}$ = a² + b² + 2ab

Then ,

-3x -2 =  x² + 2² + 2 * 2 *x

-3x -2 =  x² + 4 + 4x.

On adding both sides by 3x

-2 =  x² + 4 + 4x + 3x

-2 =  x² + 4 + 7x

On adding both sides by 2

0 =  x² + 4 + 7x + 2

On switching sides

x² +7x + 6 = 0

On Factoring

x² +6x + x + 6 = 0

x ( x+ 6 ) +1 (x +6 ) = 0

On grouping

( x +1) ( x +6) = 0

x = -1, -6.

Let check for x = -6

$\sqrt{-3 (-6) -2} = -6 + 2$.

4 =  -4

An extraneous solution is a root of a transformed equation that is not a root of the original equation.

Therefore,  -6 is the extraneous solution.