Question

Please help, been stuck on this and need a good grade, super confused. Please
When solving a radical equation, Amari and Justin came to two different conclusions. Amari found a solution, while Justin's solution did not work in the equation.

Create and justify two situations: one situation where Amari is correct and a separate situation where Justin is correct.

Answer 1

@ranga

Answer 2

Not sure what they are expecting. Is this about extraneous solutions when solving radical equations?

Answer 3

I think so, I think they're saying that Amari had a solution and that Justin didn't. so we would need extraneous and non-extraneous

Answer 4

Solve: \[\sqrt{x-2} = 5\]
Square both sides:
x - 2 = 25
x = 25 + 2 = 27

Put x = 27 back in the original equation and see if the solution is correct:
sqrt(27-2) = sqrt(25) = 5 which is the same as the right hand side and therefore the solution is correct.

Next take the radical equation: ...

Answer 5

So that could be Amari?

Answer 6

Yes. The solution worked and that could be Amari.

Answer 7

Solve: \[\sqrt{x-1} = x- 7\]
Square both sides:
x - 1 = (x-7)^2
x - 1 = x^2 - 14x + 49
x^2 - 14x - x + 49 + 1 = 0
x^2 - 15x + 50 = 0
(x-10)(x-5) = 0
x = 5 or x = 10
Put each x back in the original equation and see if they satisfy:
x = 5
LHS: sqrt(5-1) = 2
RHS: 5 - 7 = -2
LHS is not equal to RHS. So x = 5 is an extraneous solution introduced by squaring.

Try x = 10.
LHS: sqrt(10-1) = sqrt(9) = 3
RHS: 10 - 7 = 3

LHS = RHS. x = 10 is a solution.

So one of Justin's solution, x = 5, did not work.

Answer 8

thak you so much, I appreciate it I've been having a hard time!

Answer 9

You are welcome.