Question

When solving a radical equation, Beth and Kelly came to two different conclusions. Beth found a solution, while Kelly's solution did not work in the equation.

Create and justify two situations: one situation where Beth is correct and a separate situation where Kelly is correct.

Answer 1

@SolomonZelman
Could you please help me? Thanks!

Answer 2

i mean -18 and -6

Answer 3

you want a radical equation that has a solution for part A, then another equation with no solution for part B.

What the equations are is up to you.

Answer 4

@kaylak12345
Could you please post that in your own question? Thanks

Answer 5

Can I call you lover?

Answer 6

@SolomonZelman
So I need to get an extraneous solution.
And sure, I'll respond to anything haha

Answer 7

yes, extraneous solution... a solution that you arrive at at the end of the solving process but doesn't work.

Answer 8

saying doesn't work in the original equation.

Answer 9

This all is for part B. When Kelly is correct.

Answer 10

For part a, you just need a good solvable radical equation.

Answer 11

For extraneous equation, I would put something like
\(\Large\color{black}{ \frac{1}{x+2}+\frac{1}{x-2} =\frac{4}{x^2-4} }\)

Answer 12

Rather.....
\[(\sqrt{9})^2=x^2\]

Answer 13

try to solve this one, and tell me what your solution is. Then check your answer and tell me if it worked.

Answer 14

Ohh, radical equation... sorry.

Answer 15

No! It's alright! So, do I try that equation or the one you used?

Answer 16

No it is not a radical equation... it is irrelevant. although it does have an extraneous solution.

Answer 17

Alright, Thank you so much. I have all day so please do not feel rushed.

Answer 18

can one solution be extraneous, and another be not extraneous?

Answer 19

I mean 1 extraneous and another that works. can you use that?

Answer 20

I believe so

Answer 21

\(\Large\color{black}{ -\sqrt[]{x-1}=x-7 }\)

Answer 22

try to solve this one.

Answer 23

(start from raising both sides to the second power. saying- SQUARE BOTH SIDES)

Answer 24

\[x-1=x^2-49\]?

Answer 25

No exactly.

Answer 26

you got the left side correctly. and you squared the minus as well, so it is positive x-1 as you wrote.
Left side is good.\

When you square the right side, you are taking a square (second power) of \(\Large\color{black}{ (x-7) }\)
saying that this is what you do, \(\Large\color{black}{ (x-7) ^{\color{red}{2}}}\) .

Answer 27

what you wrote for the right side is missing the middle term.

Answer 28

Ah
\[x^2-14x-49\]

Answer 29

I'll keep working. Thanks for your help!

Answer 30

yes

Answer 31

So you get
\(\Large\color{blue}{ x-1=x^2-14x-49 }\)

Answer 32

Yes, so now I solve this equation?

Answer 33

yes, I would just add like terms...

Answer 34

-x on both sides,
+1 on both sides

and on....

Answer 35

-x on both sides?

Answer 36

I'm very stuck and confused now. I have no idea what I'm doing.

Answer 37

\(\Large\color{blue}{x-1=x^2-14x-49 \color{red}{ } }\)

\(\Large\color{blue}{x-1\color{red}{-x}=x^2-14x-49 \color{red}{-x} }\)

\(\Large\color{blue}{-1=x^2-13x-49 }\)

\(\Large\color{blue}{-1\color{red}{+1}=x^2-13x-49\color{red}{+1} }\)

Answer 38

it should be -15x on the left, because I subtract -14x -x

Answer 39

Oh good. I got that when I finished solving it. It just looked so weird I thought I was wrong.

Answer 40

\(\Large\color{black}{x^2-15x-48=0 }\)

Answer 41

Okay, and what are you solutions?

Answer 42

I am wrong again.

Answer 43

it was positive 49, not negatgive, so wen I add 1 to both sides, I get,
\(\Large\color{black}{x^2-15x+50=0 }\)

Answer 44

I am so confusing.