How do you solve these two extraneous solutions?

Question

4 - the square root of 3x +1 = 5

AND

the square root of 2/3 + x = - the square root of 2x+1/3

Can you please show that steps?
Thanks

Answer 1

1)
4-sqrt(3x+1)-4=5-4 subtract four from each side to isolate radical
(-sqrt(3x+1))^2=(1)^2 square each side to get rid of radical
3x+1-1=1-1 subtract 1 from each side to isolate variable
3x/3=0/3 divide each side with 2 to have x alone
x=0



2)
(sqrt(2/3+x))^2= (-sqrt(2x+1/3))^2 square each side to get rid of radicals
2/3+x-1/3= 2x+1/3-1/3 subtract 1/3 from each side to isolate variable
1/3+x-x=2x-x subtract x from each side to bring variables to one side
1/3=x

Answer 2

First, you have to get everything on one side. So subratct the four leaving you with the negative square root of 3x+1 is equal to 1. Then, square each side to get rid of the square root (fortunately, the square root of one is one!), then you can solve for x being left with 3x+1=1. Subtract the one from each side, then divide by 3 and you are left with x=0/3, or just 0. To test that simply replace x with 0 in your original formula and it works!

In the second equation start off by squaring each side to eliminate square roots, then solve for x by subtracting 1/3 from each side and x from each side should leave you with x=1/3! Again, replace x with 1/3 and it works, proving the solution.

Answer 3

4) first 3x+1>=0 so x>=-1/3
Square both sides
3x+1=25 so x=8
AND
Impossible.The left side is positive while the right side is negative.So NO solutions