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

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 $\frac{1}{2}$.
So, $\sqrt{16} = \sqrt{{4}^{2}} = {\left({4}^{2}\right)}^{\frac{1}{2}}$ and when you have an exponent to an exponent you multiply the exponents so ${\left({4}^{2}\right)}^{\frac{1}{2}} = {4}^{1} = 4$

${w}^{2} = 4$
Square root both sides to get x alone:
$\sqrt{{w}^{2}} = \sqrt{4}$
$w = \pm 2$

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. ${\left(- 2\right)}^{2} = \left(- 2\right) \cdot \left(- 2\right) = 4$