Question

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. Use a constant in place of each variable a, b, c, and d. You can use positive and negative constants in your equation.

that has an extraneous solution

Answer 1

Please I need help I only have limited time

Answer 2

The question mentions an equation that you should use as a model. May we see this equation? It seems like an important part of the question.

Answer 3

They don't give a model which is weird but you can make your own

Answer 4

Ah. Interesting.
In general, an "extraneous solution" exists when an equation can be manipulated into a new form, which produces a solution that does not satisfy the original.
That's a bit abstract, but there is a more concrete example here that should help you understand: https://www.onlinemath4all.com/extraneous-solutions-of-radical-equations.html#:~:text=Extraneous%20solutions%20of%20radical%20equations%20%3A%20Squaring%20each,the%20original%20equation%20when%20you%20solve%20radical%20equations.

Does the question perhaps give you a template that you're supposed to make your equations off of?

Answer 5

√(x-2) = (x- 4) something like this

Answer 6

Im in algebra 1

Answer 7

Interesting. I'm not sure if it's a coincidence, but the article I linked actually gives an example of how √(x-2) = (x- 4) can produce an extraneous solution.
Basically, √(x-2) = (x- 4) can be converted to
x - 2 = x^2 - 8 x + 16
by squaring both sides of the original. This new equation can be satisfied by 6 and 3, but only 6 satisfies the original, which makes 3 an extraneous solution.

So if you have that, it seems like you're halfway done! Now we just need to write an equation that does *not* have any extraneous solutions.

Answer 8

I was wondering if I could just get the answers if your okay with that. I have limited time and barley understand the work!

Answer 9

hey!

Answer 10

I think, in general, if you write a radical equation that does not include negative roots, you should be able to avoid extraneous solutions?

Answer 11

Like the example above, √(x-2) = (x- 4) had the possibility of negative roots, in case x was less than 2? So, if you write a radical equation where that doesn't happen I'm guessing you should be good.
Wanna give it a try, and we can see how it goes?

Answer 12

yeah we could do that. Im not top good at math so please bare with me

Answer 13

No problem, I'm no expert at this either, but we can try to work through it together

Answer 14

so how would you recomend we start

Answer 15

I guess one way we could start is by writing an equation that is true, like
9 = 9
Then we could think of a way to get a radical equation with variables out of that.
So, let's say we have a variable x, which is our solution. I'll make it 7, for no particular reason. In that case, this is true:
9 = x + 2
As well as 9 = √(12x + 9)
So maybe our equation could be
√(12x + 9) = (x + 2)
?

Answer 16

first of all, how familiar is the user with extraneous solies?

for extraneous solie you could use \( \sqrt{x+2}=(0-3) \), where on the other hand for a not extraneos solie, just use a number for 0, just like \( \sqrt{x+2}=(8-3) \)

work the examples, and check -- its more of a fixed thing, rather than one you can come up in the spot.

Answer 17

Well, we can square both sides and see if an extraneous solution results...
So, from that original equation
√(12x + 9) = (x + 2)
both sides get squared to become
12x + 9 = x^2 + 4x + 4
Which we can simplify into
0 = x^2 - 8x -5
.....
It's already kind of looking like this will have more than one solution. Maybe I'm going about this wrong? I get the feeling that the solution has something to do with writing an equation that is a perfect square, since there will only be one solution to that. So perhaps we should start with that and move backwards?