How do you solve #sqrt(x+1) = 8#?

1 Answer
Apr 10, 2016

Answer:

x=63

Explanation:

#sqrt(x+1)=8#
Now you have to square both sides to cencle out root.

#(sqrt(x+1))^2=8^2#
#x+1=64#
#x=64-1#
#x=63#

In order to verify: #sqrt(63+1)=8#
#sqrt64=8#
#+-8=8#
In the verification #+-# is just because of the rule any number squared gives a positive value. Accordingly #(-8)^2=64# as well as #8^2=64#.

The previous answer was not correct.