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

Answer 1

Eli Assouline

#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

Taking the 32 as an example, we want to find a squared number that 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.

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Answer 2

Eli Assouline

To solve (x^2 = 32), you would take the square root of both sides of the equation. This gives you two solutions: (x = \sqrt{32}) and (x = -\sqrt{32}). Simplifying, you get (x = \pm 4\sqrt{2}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.