Question

Please help with Algebra will award medal

Answer 1

Task 2

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

An extraneous solution is a fake result. You can go through and solve the equation, but when you check the answer it won't work.

The easiest thing to do for both equations is to make a = 1 so you don't have to worry about division. Then you're left with \(\sqrt{x+b}+c=d\)

Subtract c from both sides to get \(\sqrt{x+b}=d-c\)

For the extraneous solution equation, you need d-c to be negative. For the other one d-c will be positive.

Answer 3

so I can choose any number or does It have to be specific

Answer 4

You can choose any number

Answer 5

ok

Answer 6

so I plug in the numbers ad solve

Answer 7

I couldn't see anything in that file. Can you post a screenshot or use the draw tool?

Answer 8

ok I plugged In numbers a radical sign 18 +24+32=46

Answer 9

so
I subtract the 32 from both sides now right

Answer 10

you're still supposed to have a radical sign and an x. The only numbers you're replacing are a,b,c, and d. For example
√(x-3)+4=20

Answer 11

For that equation a =1, b=-3, c = 4, and d = 20.

Check d - c:
d - c = 20 - 4 = 16
This is positive so that's an equation without an extraneous solution

Answer 12

So I don't need to add a number for x

Answer 13

right. x stays

Answer 14

So so I subtract 4 from both sides and then now I have 1\ x+3=16 sorry I don't have a radical sign on my keyboard lol

Answer 15

So I keep simplifying the equation from there

Answer 16

yes after you subtract 4 square both sides

Answer 17

Ok now I have 1/ x + 3 =16

Answer 18

Than I subtracted 3 from both sides and now it's just 1/x=13

Answer 19

no, square both sides first then subtract 3. You can't subtract the 3 while it's under the radical.