Question

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

Answer 1

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`