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

- 6 is the extraneous solution.

Explanation:

Given :
`√(-3x -2) = x + 2`.

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

Solution : We have given that
`√(-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

`√(-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.