Question

a bit of help hard question...

Answer 1

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.

a√x+b+c=d

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

Part 2. Show your work in solving the equation. Include the work to check your solution and show that your solution is extraneous.

Part 3. Explain why the first equation has an extraneous solution and the second does not.

Answer 2

Is b inside the radical? Both b and c inside the radical?

Answer 3

b is inside the radical only until then

Answer 4

Are you sure either c or d is not multiplied by x?

Answer 5

here this is the exact thing http://prntscr.com/404e3u

Answer 6

\(\Large a\sqrt{x+b}+c = d\)
Let \(\Large a = 2, b = 3, c = -1, d = 7\)
\(\Large 2\sqrt{x+3}-1 = 7\)
\(\Large 2\sqrt{x+3} = 8\)
\(\Large \sqrt{x+3} = 4\)
\(\Large x+3 = 4^2 = 16\)
\(\Large x =13\)

No extraneous solution.

Answer 7

i need two equtions tho.... so you would use the same letters for numbers?

Answer 8

Yeah, I created the second equation with no extraneous solutions. But they want the first equation to have extraneous solutions.
Not sure if a radical equation with extraneous solution can be created in the model of \(\Large a\sqrt{x+b}+c = d\) by just choosing a set of numbers for a,b,c and d.

Answer 9

oh ok but how should i make the second one?

Answer 10

But if you are allowed to create ANY radical equation that has an extraneous solution (not necessarily conforming to the model of \(\Large a\sqrt{x+b}+c = d\) then we can do so.

For example, \(\Large \sqrt(x-1) = x- 7\) when solved for x will have an extraneous solution.

Answer 11

My guess is, they meant "cx" in the model equation and not just "c". It may have been a typo.

Answer 12

u still need help?

Answer 13

no nawt really

Answer 14

but thx