Question

does anyone know how to do extraneous soultions for radicals?

Answer 1

To check for extraneous solutions, plug each of your solutions in for x into the radical. If you get a negative radical, the solution was introduced through squaring and can be eliminated.

Answer 2

i dont understand this at all.

Answer 3

Do you have solutions like x=0, x=1, etc.?
Take these and plug them into the original problem. If you get a negative under the radical, don't count that as an answer.

Answer 4

how do you get the solutions?

Answer 5

\[5\sqrt{x}=1\] is one of the problems but i dont know how to find the solution

Answer 6

Square it, so you get
25*x=1
So x=1/25.
Plug it back in
5*sqrt((1/25))=1, which works. 1/25 is not an extraneous solution.

Answer 7

thank you, how would i do one like this: \[(x-3)1/3=6\]

Answer 8

That's simple. Just multiply by 3 on both sides to get rid of the 1/3.
(x-3)=6*3
x-3=18
x=21

Answer 9

thank you that helped me with another one similar to that, i have one last question on how to do one like this: \[\sqrt{x}+1=\sqrt{2x}-7\]

Answer 10

OK, let's get everything organized to be solved.
8=\[\sqrt{2x}-\sqrt{x}\]
8=\[(\sqrt{2}-1)*\sqrt{x}\]
8/\[(\sqrt{2}-1)\]=sqrt(x)
square both sides. then plug in your solutions to the original equation to see if they work.