Question

I need an example of an extraneous solution.

Answer 1

@I_Always_Smiling @Taylev105

Answer 2

@pooja195 @JoeJoldin

Answer 3

please the subsequent equation:
\[\sqrt{2(x ^{2}+1)}=3x+1\]
has two solutions, namely:
\[x=\frac{ 1 }{ 7 },x=-1\]
nevertheless, the solution:
\[x=-1\]
is extraneous, why?

Answer 4

It's extraneous because it doesn't come to the same solution on both sides.

Answer 5

@Michele_Laino

Answer 6

@just_one_last_goodbye

Answer 7

correct

Answer 8

& that's all I needed. Thanks bunches. cx

Answer 9

welcome!

Answer 10

Also, (Just thought of this) do you think you can coach me into making my own equation?

Answer 11

@TuringTest

Answer 12

you mean one with an extraneous solution?

Answer 13

Right.

Answer 14

sure, just take any true equation and either multiply it by zero, or a variable:
\(x+3=0\) what is the solution?

multiply both sides by x and solve again:
what is the solution now?

Answer 15

6=0?

Answer 16

that's right! @Brostep0s
since 3*(-1)+2 is negative

Answer 17

how did you get the 6?

Answer 18

whereas the square root is positive!

Answer 19

It's wrong, the answer is 0=0.

Answer 20

yeah but I told you to multiply both sides of \(x+3=0\) by \(x\), I don't know where you got the 6 from

Answer 21

you wanted to make an equation with an extraneous solution from scratch, so that's what I'm trying to show you

Answer 22

I don't understand??

Answer 23

do you want to create an equation from scratch that has an extraneous solution?

Answer 24

That would be the goal, yes.

Answer 25

Ok, so first we just make up a super simple equation that we can easily find the answer to. Let's go with\[x+3=0\]what is the solution to this equation?

Answer 26

@TuringTest How do you square the right side of the equation @Michele_Laino gave me?

Answer 27

FOIL
(3x+1)(3x+1)

Answer 28

15x+1?