Which of the following is an extraneous solution of mc016-1.jpg? x = –6 x = –1 x = 1 x = 6

[Incomplete question. I found the equation on other source:

√(-3X-2)=X+2]

Let’s find x:

√(-3X-2)=X+2

-3X-2=(X+2)²

-3X-2=x²+2*2*x+2²

-3X-2=x²+4x+4

-3X-2-x²-4x-4=0

-x²-7x-6=0

So now you should use Bhaskara formula. Unless your calculator makes for itself (some calculators do).

ax²+bx+c=0

x=(-b+-√(b²-4ac))/(2a)

a=-1

b=-7

c=-6

x=-1

or x=-6

√(-3*(-1)-2)=-1+2

√(3-2)=1

√1=1

1=1 TRUE -> So -1 is a solution, not an extraneous solution

Let’s verify x=-6:

√(-3*(-6)-2)=-6+2

√(18-2)=-4

√16=-4

attachment?

step-by-step explanation:

maybe i can