How do you solve #x^2 = 32#?

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Answer 1 · 2018-03-22T16:01:51

#x=4sqrt(2)" "# Exact answer for principle root

#x=+-4sqrt(2)" "# As a full exact answer

#x+-5.6569 # to 4 decimal places - Approximate answer

Consider the 32. We are looking for squared number we can 'take outside' the root

32 is even so we have #2xx16#

But 16 is #4^2# giving #2 xx 4^2#. So we can write:

#x=sqrt(2xx4^2)#

#x=4sqrt(2)# Using the Principle Root which is always positive.