Question

Identify any extraneous solutions @YOLO_KING

Answer 1

Do you have choices?

Answer 2

Yep hold on a sec :)

Answer 3

\[x=\sqrt{4+12}\]
So I got x= 4

Answer 4

Because 4 + 12 = 16 and \[\sqrt{16}= 4\]
So x= 4

Answer 5

A.) 2 is a solution to the original equation. The value -6 is an extraneous solution.
B.) 6 and -2 are solutions
C.) 6 and 2 are both extraneous solutions
D.) 6 is a solution to the original equation. The value of -2 Is an extraneous solution

Answer 6

Wait im comfused so are these choices part of the x=\[\sqrt{4+12}\] problem?

Answer 7

Yep its said identify any extraneous solutions

Answer 8

hmmmmm... I am not sure about this one let me get some help

Answer 9

Okay the last question is like this to just a heads up :)

Answer 10

@campbell_st Could you help out?

Answer 11

kk

Answer 12

I have no idea about this one :/

Answer 13

well the equation would appear to be

(x -6)(x + 2) = 0

or

\[x^2 - 4x - 12 = 0\]

so is the problem

\[x = \sqrt{4x + 12}\]

Answer 14

Yes it is

Answer 15

ok... so yo solve it square both sides of the equation and you get

\[x^2 = 4x + 12\]

subtract 4x and 12 from both sides and its

\[x^2 - 4x - 12 = 0\]

this can be solved by factoring

(x + 2)(x - 6) = 0

which gives -2 and 6 as solutions

now substitute these into the original question to see if anything doesn;t work

so

\[x = \sqrt{4x + 12}~~ x = 2~~~then~~~ 2 = \sqrt{4 \times 2 + 12}\]

does that work...?

and then repeat it with x = 6

\[6 = \sqrt{4 \times 6 + 12}\] does that work...

an extraneous solution is one that may be a solution, but when its tried in the original equation it doesn't work

Answer 16

Thank you!