Question

Extraneous equations?

Answer 1

I have to create my own extraneous equation using the formula
\[x+a = \sqrt{bx+c}\]
Where a, b, and c are all positive integers and b >1. The solution has to equal 7 and there must be an extraneous solution. Do you know how I would do that?

Answer 2

Pick b(7) + c < 0. That should do it.

Answer 3

so, \[x +a = \sqrt{b(7) + c}\] correct?

Answer 4

You need to choose a, b, and c. Unfortunately, I missed that ALL of a, b, and c must be POSITIVE. We can't get b(7) + c < 0 like that.

We do need \(7 + a = \sqrt{b(7)+c}\)

Answer 5

ok, so x is 7. I just choose a, b, and c. got it. Thnx

Answer 6

Not quite. I suspect there are MANY solutions to this problem statement.

Personally, I picked a = 1 and b = 3 and solved for c. Can you imagine how to do that?

Answer 7

ok, so im thinking like you said x=7 and a=1 so
(7) + (1) =
then square it off and solve.
Then ill choose integers for b and c to make the equation true

Answer 8

Well, we DO have to consider the equation we are actually given.

With a = 1 and b = 3, we have \(x + 1 = \sqrt{3x+c}\)
Squaring \(x^{2} + 2x + 1 = 3x + c\)
Standard Form \(x^{2} - x + (1-c) = 0\)

If we have x = 7, this this thing MUST have a factor of (x-7).

"Q" is a number we don't yet know.

Factoring with Q: \((x-7)(x-Q) = x^{2} - x + (1-c) = 0\)
Expanding the Left-Most \(x^2 -7x - Qx + 7Q = x^{2} - x + (1-c) = 0\)
And this we see: 1) - 7 - Q = -1 and 2) 7*Q = 1-c

Look at each step. Consider it carefully.
Do you see how to proceed?

Answer 9

tbh im lost

Answer 10

\[x+a = \sqrt{bx+c}\]
We need \(bx+c>0\) and we need \(\sqrt{bx+c} > 7+a\) where \(a\) is a positive integer.
\[7+a = \sqrt{7b + c}\]How about \(b=10\), what would be a good value for \(c\)?

Answer 11

(sorry, I omitted the description that I was substituting \(7\) for \(x\), because we know that is the solution)

Answer 12

@royalbug

Answer 13

if b = 10 then a good value for c would probably be 11 right?

Answer 14

You can pick any two. After that, your choices for the third are substantially limited.

For example, with a = 2 and b = 3 (ignoring that x = 7 is a solution), \(c \ge 15/4\).
For example, with a = 6 and b = 10 (ignoring that x = 7 is a solution), \(c \ge 35\).

Answer 15

@royalbug you need to solve the equation again and make sure that you have an extraneous solution. I think you will find that \(b=10,c=11\) does not work in that regard, but there are other values of \(c\) which will.

Answer 16

ok, I got two solutions. Can you tell me if they are correct?
\[x+1=\sqrt{3x+43}\]
7 = true and -6 false
also,\[x+2=\sqrt{5x+46}\]
7 = true and -6 = false

Answer 17

Both of those equations appear to meet all of the requirements, good work!

Answer 18

Thanks!