#### Archie Deetoo

##### New member

- Joined
- Mar 11, 2018

- Messages
- 5

Solve the given equation:

sqr root (3x+2) = 3x

I squared both side and then set to zero and ended up with:

(3x+1)(3x-2)=0

x = -1/3, 2/3

Now because I squared both sides in solving, I need to check for extraneous roots.

I checked x = 2/3 and got:

sqr root (3 (2/3) +2)) ?=? 3(2/3)

sqr root (4) = 2 : which is correct, so that is a correct root,

Now I checked x=-1/3:

sqr root (3(-1/3) +2)) ?=? 3(-1/3)

sqr root (1) = -1. That is also correct because square root of 1 is positive 1 and negative 1. So it appears both are correct and not extraneous solutions, yet the answer is 2/3 only. Why is that?

I have graphed the equation and can see that 2/3 is the correction solution and I can see by graphing that the extraneous solution is introduced when I square both sides when solving it. It seems that square root of 1 is only positive 1 on this occasion which is why -1/3 is extraneous.

I suppose what this comes down to is why is sometimes the square root of a number a positive answer and sometimes a positive and negative answer depending on the problem?

I appreciate any help with this.

Thanks a lot.

Zen