Question

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

@ayeitsJenni. @dwatts158

Answer 2

@mathmath333

Answer 3

first thing is do u know what the radical equations look like

Answer 4

How do I find that?

Answer 5

@mathmath333

Answer 6

It's extraneous right?

Answer 7

\[\sqrt{x-2}=5\]

Answer 8

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 9

yes this is an example for radical equation

Answer 10

So I could use this?

Answer 11

What's one that won't work?

Answer 12

what is

\(\Large \sqrt{25}\)

Answer 13

5

Answer 14

what is \(\Large -5\times -5=\)

Answer 15

25?

Answer 16

yes

Answer 17

so actually

\(\huge \sqrt{25}=\pm5\)

Answer 18

did u get that

Answer 19

Yea, and?

Answer 20

so if u consider

\(\sqrt{x-2}=5\)

for which u r getting the answer as \(27\)

\( \sqrt{27-2}=\sqrt{25}\\= -5\)

if u consider the sqrare root of 25 as -5 it wull be wrong according to your original equation

cuz

\(-5\neq 5\)

Answer 21

OH!!! That makes sense... THANKS a lot!

Answer 22

yw