## anonymous 5 years ago does anyone know how to do extraneous soultions for radicals?

1. anonymous

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.

2. anonymous

i dont understand this at all.

3. anonymous

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.

4. anonymous

how do you get the solutions?

5. anonymous

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

6. anonymous

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.

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

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.