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

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0

for this case we have by definition, if two lines are perpendicular, then the product of their slopes is -1.

then, given the following line:

so,

we are looking for m_ {2}:

then, the line is given by:

we find "b" replacing the given point:

finally, the equation is:

option c