Question

Algebra 1 question
Medal + Fan

Answer 1

Using the equation below as a model, fill in numbers in the place of a and b to create a rational equation that has an extraneous solution.

x plus a over ax = b over x

Part 1. Show all work to solve for x in the equation and check the solution.

Part 2. Explain how to identify the extraneous solution and what it means.

Answer 2

Can you help @surjithayer ?

Answer 3

is this your statement?
\[x+\frac{ a }{ ax }=\frac{ b }{ x }\]

Answer 4

not quite... ill post a photo

Answer 5

If I choose a=1 and b=2, would it still have a extraneous solution?

Answer 6

Are you able to solve for x without having to replace 'a' and 'b' with numbers?

Answer 7

Im not supposed to, but I guess I probably could if needed...

Answer 8

That's what I recommend. Tell me what you get (you should get 2 solutions).

Answer 9

hmm.... I got b=((a+x)/(b)) and a=((−x)/(−b+1))

Answer 10

you need to solve for x

Answer 11

ok then, ax+x^2=abx and x=ab-a

Answer 12

I was thinking more along like this...

(x+a)/(ax) = b/x
x(x+a) = ax*b ... cross multiply
x(x+a) - ax*b = 0
x(x+a - ab) = 0 ... factor
x = 0 or x+a - ab = 0
x = 0 or x+a = ab
x = 0 or x = ab-a

Answer 13

Notice how no matter what 'a' and 'b' are, the possible solution x = 0 pops up.

However, we must check every possible solution back into the original equation.

(x+a)/(ax) = b/x

(0+a)/(a*0) = b/0 ... and we run into a division by zero error here

so x = 0 is an extraneous solution

Answer 14

\[x^2+ax=abx\]
\[x^2+ax-abx=0\]
\[x^2-(ab-a)x=0\]
\[x[x-a(b-1)]=0\]
either x=0
but \[x \neq 0\]
therefore x is an extraneous root,a & b can have any value.
or x=a(b-1)

Answer 15

x = ab-a is the only possible solution left, but it cannot be zero or else you'll run into the same issues above


x = ab-a
x = a(b-1)

if x = 0, then a(b-1) = 0 which leads to a = 0 or b-1 = 0 --> b = 1

So if a = 0 or b = 1 (or both), then x = ab-a will be zero. That means we will have another extraneous solution. If 'a' is not zero, and b is not 1, then you'll have no division by zero issues to worry about.

Answer 16

Ok.... data overload :0.... can you do a short recap of what you just said?

Answer 17

I solved for x in general (in terms of 'a' and 'b'). When I did that I got these 2 possible solutions

x = 0 or x = ab - a

The first solution, x = 0, doesn't work in the original equation. So it's extraneous

Answer 18

here for x to be extraneous root
\[a \neq0,b \neq 1\]

Answer 19

the only other solution x = ab-a can be any number but zero

to make sure ab-a is nonzero, we must make sure that 'a' is not zero and b is not 1

Answer 20

thank you...

Answer 21

Are we done with part A?

Answer 22

well I haven't done part A or part B. I just solved it in general to show you that no matter what numbers you pick for 'a' and 'b', you'll always get at least 1 extraneous solution

Answer 23

you said "If I choose a=1 and b=2, would it still have a extraneous solution?"

the answer is "yes, because x = 0 is always one of the possible solutions, so x = 0 is extraneous" and you would show why x = 0 is extraneous

Answer 24

Alright 'o, thanks for clarifying that :D

Answer 25

with a = 1 and b = 2, this means


x = ab-a

x = 1*2-1 ... plug in a = 1 and b = 2

x = 2-1

x = 1

So if a = 1 and b = 2, then x = 1 is the solution (turns out the only solution)

Answer 26

or you could pick a = 1 and b = 2, plug them into the original equation, then solve for x

Answer 27

Thanks a ton, now were off to B.... right?

Answer 28

yeah you just need to explain how to ID the extraneous solutions

Answer 29

ID... like identify?

Answer 30

yeah, sry I was being lazy with typing lol

Answer 31

no no thats ok.... would you just simply say:

Once you solve the equation you're given, take the solutions and enter them into the original equation. If the answer comes out to be equal to what its supposed to be equal to, then the solution is valid. If the answer is not equal to what it was supposed to be equal to, then the solution is extraneous.

Answer 32

yes, also tack on "if plugging a possible solution back into the original problem causes a division by zero error, then that possible solution isn't really a true solution at all. It's extraneous" or something like that


It turns out that it all has to do with domain. If you get a possible solution that is not in the domain, then it's not a true solution. This solution is extraneous.

Answer 33

OK... did that :) Time for part 2

Answer 34

I thought that was part 2?

Answer 35

part 2 is " Explain how to identify the extraneous solution and what it means."

Answer 36

OOOPS OMG I MESSED UP THANKS A TONGAZILION :D

Answer 37

Can you help me on another question?

Answer 38

sure, what is it?

Answer 39

Create a quadratic polynomial function f(x) and a linear binomial in the form (x − a).

Part 1. Show all work using long division to divide your polynomial by the binomial.

Part 2. Show all work to evaluate f(a) using the function you created.

Part 3. Use complete sentences to explain how the remainder theorem is used to determine whether your linear binomial is a factor of your polynomial function

Answer 40

alright, go ahead and post what you have so far, I'll brb in a few min

Answer 41

ok back