Question

@perl Part 1. Create two radical equations: one that has an extraneous solution, and one that does not have an extraneous solution. Use the equation below as a model.
a√x+b+c=d
Use a constant in place of each variable a, b, c, and d. You can use positive and negative constants in your equation.
Part 2. Show your work in solving the equation. Include the work to check your solution and show that your solution is extraneous.
Part 3. Explain why the first equation has an extraneous solution and the second does not.

Answer 1

@perl There is three questions on this test that i am stuck on and this is the 2nd one

Answer 2

btw its

Answer 3

@itsbribro @AlexandervonHumboldt2

Answer 4

@dtuniyants

Answer 5

@Catlover5925

Answer 6

try one with d = 0, and pick c so that it is negative

Answer 7

so it would be a square root x+b+(-1)=0?

Answer 8

my idea was
1*sqrt(x + 2) -1 = -1

if you add 1 to both sides you get

1*sqrt( x+2) = 0

that has only one solution if you square both sides
(x+2) = 0
x = -2

Answer 9

this equation does not have an extraneous solution

Answer 10

dang your fast.

Answer 11

it shows show why it is extraneous

Answer 12

actually any positive value will work

Answer 13

1*sqrt(x + 2) -1 = 3

works as well, there is no extraneous solution

but if you do

1*sqrt(x+2) -1 = -3 , you get an extraneous solution. do you see why?

Answer 14

try to solve
1*sqrt(x+2) - 1 = -3

and then plug in the solution

Answer 15

after i add one do i square root both sides?

Answer 16

you can square both sides

Answer 17

when i square root -2 i get an error message

Answer 18

you should square both sides , not square root

Answer 19

1*sqrt(x+2) - 1 = -3
sqrt(x+2) = -2

we want to solve for x, so we square both sides

sqrt( x+2)^2 = (-2)^2

x+ 2= 4

Answer 20

oh okay i got it now

Answer 21

now if you plug in that solution back into the original equation, what happens?

Answer 22

hold on what is an extranous solution?

Answer 23

It equals itself right? @perl

Answer 24

nope

Answer 25

x would equal 2 right?

Answer 26

an extraneous solution is like an illegal solution , it comes extra when you solve for x

Answer 27

lets plug in x = 2 into the original equation 1*sqrt(x+2) - 1 = -3

1*sqrt( 2 + 2 ) -2 = sqrt(4) -2 = 2 - 2 = 0
but 0 does not equal to -3

Answer 28

okay

Answer 29

so it turns out x = 2 is an invalid solution, so we call it extraneous . we got it when we tried to solve for x by squaring

Answer 30

in general, when you square an equation, you can introduce an invalid or extraneous solution

Answer 31

so that is what i would put for part 3 and then my work for part 2?

Answer 32

so for the first one that is not extranous what happens when you plug in -2?

Answer 33

but the equation
1*sqrt(x + 2) -1 = 3

this does not have an extraneous solution. when we solve it we get

sqrt(x+2) = 4
(sqrt(x+2))^2 = 4^2
(x+2) = 16
x = 14

if you plug in x = 14 into
original equation, it satisfies it

Answer 34

you said the equation was 1*sqrt(x+2)-1=-1

Answer 35

yes i kind of changed it :0

Answer 36

that equation works as well
if you plug in x = -2 into the equation 1*sqrt(x +2) -1 = -1

you get
1*sqrt( -2 + 2 ) -1 = sqrt(0) -1 = 0 -1 = -1

so it works and x=-2 is not an extraneous solution

Answer 37

okay i got one last question that i tagged you in!