# Square roots

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.

Posted on Categories Algebra, Pre-VCE