Square roots

Multiplying square roots

The key fact of square roots is that they can be multiplied. For example, we can do

$\sqrt{9} \cdot \sqrt{4} = \sqrt{36}$

We know this works because

$$\sqrt{9}=3 , \sqrt{4}= 2 , \text{and} \sqrt{36} = 6$

$\sqrt{9} \cdot \sqrt{4} = 3 \cdot 2 =6 =\sqrt{36}$

This also works for numbers that don’t have whole number square roots. For example, we can say

$\sqrt{5} \cdot \sqrt{20} = \sqrt{100}$

even though 5 and 20 don’t have simple square roots.
We can check that this makes sense with a calculator. We know $\sqrt{100} = 10$ , and on a calculator we can get $\sqrt{5} \approx 2.236$ and $\sqrt{20} \approx 4.472$ . Now on a calculator we get

$\sqrt{5} \cdot \sqrt{20} \approx 2.236 \cdot 4.472 \approx 9.99$

so we’re pretty close.

You can divide square roots similarly. For example, $\sqrt{20} \div \sqrt{4} = \sqrt{5}$ .

Simplifying square roots

Since we can multiply square roots, we can also simplify them by breaking them up. Let’s try an example: simplifying $\sqrt{18}$ . First we look for a perfect square that goes into 18. 9 goes into 18! So we can break up $\sqrt{18}$ into $\sqrt{9} \cdot \sqrt{2}$ because

$\sqrt{18} = \sqrt{9} \cdot \sqrt{2}$

But

$\sqrt{9} = 3$

$\sqrt{18} = 3 \cdot \sqrt{2}$

$3\sqrt{2}$ is the fully simplified form of $\sqrt{18}$ because the $\sqrt{2}$ can’t be simplified any further; we’ve taken as much out of the square root as we can.

Multiplying square roots by themselves

Often on the SHSAT, you see a square root multiplied by itself, like $\sqrt{10} \cdot \sqrt{10}$ . Since we know square roots can be multiplied,

$\sqrt{10} \cdot \sqrt{10} = \sqrt{100} = 10$

Is it just a coincedence that $\sqrt{10} \cdot \sqrt{10} = 10$ ? No! In fact, multiplying the square root of a positive number by itself always gives you the number! This is actually basically the definition of a square root. A few examples:

$\sqrt{5} \cdot \sqrt{5} = \sqrt{25} = 5$

$\sqrt{16} \cdot \sqrt{16} = 4 \cdot 4 = 16$

$\sqrt{37} \cdot \sqrt{37} = 37$

This trick can save a lot of time, especially with bigger numbers.

Square roots and geometry

So what kind of square root problems might we see on the test? A lot of the time when you see square roots, they are in geometry problems. A couple examples:
q) What is the area of the figure?

Solution

q) An isosceles right triangle has a hypotenuse that is 8 units long. What is its area?

Solution

Estimating square roots

It is easy to calculate the square roots of perfect squares, like 16. We know $\sqrt{16} = 4$ . Other numbers, though, we need a calculator for. For example, we can put $\sqrt{23}$ into a calculator and we will get 4.795831523, which is actually just part of an infinitely long decimal. You won’t have to get exact values for these types of square roots on the SHSAT, but you should be able to estimate their value.

How do we estimate square roots? It’s simple! Let’s look at an example, say $\sqrt{32}$ . 32 is between the perfect squares 25 and 36, so $\sqrt{32}$ is between $\sqrt{25}$ and $\sqrt{36}$ . So $\sqrt{32}$ is between 5 and 6.

Now we can answer questions like

q) Which is bigger, $3 + \sqrt{15}$ or $5+\sqrt{6}$ ?

Solution

Warning: adding square roots

If you have to add square roots, be careful! You can’t add square roots as simply as you can multiply them. Let’s try an example:

$\sqrt{16}+\sqrt{9}=4+3=7\not=\sqrt{25}$

So we see $\sqrt{16}+\sqrt{9}\not=\sqrt{25}$ .

The only way you can really add square roots is by combining common factors, in the same way that you can do $3x + 6x = 9x$ . You can only add multiples of the same square root