Respuesta :
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
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:
- 6 is the extraneous solution.
Step-by-step explanation:
Given : [tex]\sqrt{-3x -2} = x + 2[/tex].
To find : Which of the following is an extraneous solution .
Solution : We have given that [tex]\sqrt{-3x -2} = x + 2[/tex].
Taking square both sides
-3x - 2 = [tex](x+2)^{2}[/tex].
On applying identity [tex](a+b)^{2}[/tex] = 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
[tex]\sqrt{-3 (-6) -2} = -6 + 2[/tex].
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.
