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What is the square root of 90 simplified in radical form?

Jul 13, 2016

$\sqrt{90} = 3 \sqrt{10}$

Explanation:

To simplify $\sqrt{90}$, the goal is to find numbers whose product gives the result of $90$, as well as collect pairs of numbers to form our simplified radical form.

In our case, we can begin in the following way:

$90 \to \left(30 \cdot 3\right)$

$30 \to \left(10 \cdot 3\right)$ ... $\cdot$... $3$

$10 \to \left(5 \cdot 2\right)$ ...... $\cdot$... ${\underbrace{3 \cdot 3}}_{p a i r}$

Since we don't have numbers we could further divide which yield a number other than $1$, we stop here and collect our numbers.

A pair of numbers counts as one number, namely the $3$ itself.

Thus we can now write $\sqrt{90} = 3 \sqrt{5 \cdot 2} = 3 \sqrt{10}$

More examples:

(1) $\sqrt{30}$

$30 \to \left(10 \cdot 3\right)$
$10 \to \left(5 \cdot 2\right)$ ... $\cdot$... $3$

We cannot find any more divisible factors, and we certainly don't have a pair of numbers, so we stop here and call it not simplify-able. The one and only answer is $\sqrt{30}$.

(2) $\sqrt{20}$

$20 \to \left(10 \cdot 2\right)$
$10 \to \left(5\right) \cdot {\underbrace{2 \cdot 2}}_{p a i r}$

We've found a pair, so we can simplify this one:

$\sqrt{20} = 2 \sqrt{5}$

(3) $\sqrt{56}$

$56 \to 8 \cdot 7$
$8 \to 4 \cdot 2 \cdot 7$
$4 \to {\underbrace{2 \cdot 2}}_{p a i r} \cdot 2 \cdot 7$

We proceed the same way and write $\sqrt{56} = 2 \sqrt{2 \cdot 7} = 2 \sqrt{14}$