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

1 Answer
Sally
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

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