Square Roots


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Estimate Square Roots

Now because we are familiar with 1^2=1,\,2^2=4,\,3^2=9,\,\dots 12^2=144, we can figure the two integers a square root (from zero to 144) lies between.

For example, \sqrt{30} lies between 5 and 6, because 5^2=25 and 6^2=36.

Simplifying a surd

Often times a non-square number under a square root (radicand) can be written as a product of a square number by a non-square number. This is called simplifying the surd.

First, you need to factor the radicand. We’re looking for the highest factor that is a perfect square.

The first 12 positive square numbers are:

    \[ \boxed{1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144} \]

In the applet below, think of the highest square number that will divide (is a factor of) the integer given.

Example: Simplify \sqrt{72}

Let’s list the factors of 72.

Here are the factors: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72

The factors that are also perfect squares are: 4, 9, 36 (also 1 but 1 is not useful here).

The highest is 36.

So, \sqrt{72}=\sqrt{36\times 2}=\sqrt{36}\times \sqrt{2}=6\sqrt{2}

Suppose I selected 9 or 4 because I didn’t spot that 36 was a factor. In this case, I end up with another surd that can be simplified, so I just simplify again.

e.g. \sqrt{72}=\sqrt{9\times 8}=\sqrt{9}\times \sqrt{8}=3\sqrt{8}

and \sqrt{8}=\sqrt{4 \times 2}=\sqrt{4} \times \sqrt{2}=2\sqrt{2}

Therefore, \sqrt{72}=3\sqrt{8}=3\times2 \sqrt{2}=6\sqrt{2}

Simplify the square roots by writing as a product of a square number with a non square number:

Rationalise a denominator

It is considered ‘more simple’ to multiply by an irrational number than to divide by one. Suppose we have the fraction:

\frac{5}{\sqrt{3}}

We can rationalise the denominator by multiplying this fraction by 1, where 1=\frac{\sqrt{3}}{\sqrt{3}} as follows:

 \frac{5}{\sqrt{3}}=\frac{5}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{5\sqrt{3}}{3}

We end up with a rational number on the denominator, which was the goal. This is often used to simplify lines of work in trigonometry.

\sin(45°)=\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}

Here’s another trick of numbers to simplify an irrational denominator. Suppose we have the fraction

\frac{5}{11+\sqrt{3}}

We can rationalise this denominator by multiplying this fraction by 1, where 1=\frac{11-\sqrt{3}}{11-\sqrt{3}} as follows:

    \begin{align*}&\frac{5}{11+\sqrt{3}}\\[10pt]&=\frac{5}{11+\sqrt{3}}\times \frac{11-\sqrt{3}}{11-\sqrt{3}}\\[10pt] &=\frac{5(11-\sqrt{3})}{(11+\sqrt{3})(11-\sqrt{3}})\\[10pt] &= \frac{5(11-\sqrt{3})}{11-\sqrt{3}+\sqrt{3}-9}\\[10pt]&=\frac{5(11-\sqrt{3})}{2} \end{align*}

Again, our denominator is now rational.

Try the ten questions at the bottom of this mathisfun page!


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