Question

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

Answer 1

[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

Let’s verify -1

√(-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

4=-4 FALSE –> So -6 comes from the equation but does not verify. So -6 is an extraneous solution. This is related to the fact that square roots have only 1 solution, whereas squares have 2 solutions (remember (-1)² is 1, and 1² is 1 as well, because of the rules of signs for multiplication).

Answer 2

Answer:the answer is x=-1,x=7

Step-by-step explanation: