Anyone Really good at Solving Radical Equations?

Question

anonymous · Answer

I guess i can give it a shot . Let's go !

anonymous · Answer

alright then

anonymous · Answer

I'm looking for "x" right ?

anonymous · Answer

yeah and identify any extraneous solutions, it would also help if you showed your work.

anonymous · Answer

the first 1 is √8 +1 not 11

anonymous · Answer

Well , for the first one the answer is
x=3 
8*3+1 = 25
And the square root of 25 is 5

anonymous · Answer

$$\sqrt{x-7}+5=11$$ subtract 5$$\sqrt{x-7}=6$$ square both sides 
$$x-7=36$$ add 7$$x=43$$

anonymous · Answer

$$-4\sqrt{x+9}=20$$ has no solution because the left hand side is negative and  the right hand side is positive

anonymous · Answer

$$\sqrt{6x+5}=\sqrt{53}$$ square both sides and solve 
$$6x+5=53$$

anonymous · Answer

Wow dude, you just blew me away.

anonymous · Answer

For the last one , 
x=3
2+4=6 
3+6 = 9
The square root of 9 is 3

anonymous · Answer

i actually understand this now, thanks so much you guys.