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

b. an equilateral triangle inscribed in a circle

step-by-step explanation:

we are given the figure of two circles having a triangle inside.

now, we know that,

"when a figure is lying inside an another figure, then the outside figure is known as circumscribed figure and the figure lying inside is referred to as the inscribed figure".

as we see that, the circles are containing the triangle completely.

thus, the circles are the circumscribed figures and the triangle is the inscribed figure.

hence, we get, 'an equilateral triangle is inscribed in a circle'.