How do you solve #w^2 = 4# where w is a real number?

1 Answer
Apr 8, 2015

To solve this, you have to think about what get rids of a square root.

Think about the number 16...#16=4^2#. But #sqrt(16)=4#. Can you see how the square root gets rid of the square exponent?

Another way to think about it this is that taking a sqare root of something means the same thing as taking that number to the power of #1/2#.
So, #sqrt(16)=sqrt(4^2)=(4^2)^(1/2)# and when you have an exponent to an exponent you multiply the exponents so #(4^2)^(1/2)=4^1=4#

#w^2=4#
Square root both sides to get x alone:
#sqrt(w^2)=sqrt(4)#
#w=+-2#

Notice that w can equal two real numbers: +2 and -2. The positive root is quite intuitive- But the negative root also holds true, because when you mutiply two negative numbers, you get a positive number. #(-2)^2=(-2)*(-2)=4#