Now eventually, I would like to be able to solve equations like \[

{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{7}\hspace{0.33em}{=}\hspace{0.33em}{18}

\]. But you see that if I am to get *x* by itself, I need something to get rid of the exponent “2”. This is where *square roots* come in. But let’s first talk about *inverse operations.*

You may have already noticed that addition and subtraction are inverse operations because if you add a number to something and then subtract it or vice versa, you are left with the original something:

*x* + 3 – 3 = *x*, *x – 3 + 3 = *x

The same can be said of multiplication and division:

\[{3}{x}\hspace{0.33em}\div\hspace{0.33em}{3}\hspace{0.33em}{=}\hspace{0.33em}\frac{\rlap{/}{3}x}{\rlap{/}{3}}\hspace{0.33em}{=}\hspace{0.33em}{x}

\]

The square root of a number is the inverse of squaring a number. Where squaring a number means “what do I get when I multiply a number by itself”, the square root of a number means “what number multiplied by itself is equal to the original number”. The math symbol which means “take the square root of this number” is called a radical and looks like\[

\sqrt{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}

\]. You will see later that there are other ways to indicate a square root. Now a few examples:

\begin{array}{l}

{\sqrt{25}\hspace{0.33em}{=}\hspace{0.33em}{5}{,}\hspace{0.33em}{\mathrm{because}}\hspace{0.33em}{5}^{2}\hspace{0.33em}{=}\hspace{0.33em}{25}}\\

{\sqrt{4}\hspace{0.33em}{=}\hspace{0.33em}{2}{,}\hspace{0.33em}{\mathrm{because}}\hspace{0.33em}{2}^{2}\hspace{0.33em}{=}\hspace{0.33em}{4}}\\

{\sqrt{100}\hspace{0.33em}{=}\hspace{0.33em}{10}{,}\hspace{0.33em}{\mathrm{because}}\hspace{0.33em}{10}^{2}\hspace{0.33em}{=}\hspace{0.33em}{100}}

\end{array}

\]

So let’s illustrate the “inverseness” of this:

\[\begin{array}{l}

{\sqrt{{5}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{5}{,}\hspace{0.33em}\hspace{0.33em}{\left({\sqrt{5}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{5}}\\

{\sqrt{{2}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{2}{,}\hspace{0.33em}\hspace{0.33em}{\left({\sqrt{2}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{2}}\\

{\sqrt{{10}^{2}}\hspace{0.33em}{=}\hspace{0.33em}{10}{,}\hspace{0.33em}\hspace{0.33em}{\left({\sqrt{10}}\right)}^{2}\hspace{0.33em}{=}\hspace{0.33em}{10}}

\end{array}

\]

Are you starting to see the potential of this to get rid of the “2” in \[

{x}^{2}

\]?

Now I’ve only been half truthful so far. For the square root examples I’ve shown you, I only showed you one of the possible solutions. Is 5 the only square root of 25? Think about it. Is there not another number that when multiplied by itself is 25? What about -5? When you take the square root of a number, you actually are introducing two solutions:

\[\sqrt{25}\hspace{0.33em}{=}\hspace{0.33em}\pm\hspace{0.33em}{5}

\]

The symbol ± means “plus or minus” and indicates that the two solutions are +5 and -5.

Let’s use this new tool to solve an equation on my next post.