# What is the sqrt145 in simplest radical form?

Dec 19, 2017

$\setminus \sqrt{145} = \setminus \sqrt{5 \cdot 29}$

5 and 29 are both prime numbers , so the simplest form of $\setminus \sqrt{145}$ is $\setminus \sqrt{145}$

Dec 19, 2017

Simplest form = $\sqrt{145}$

$\approx 12.042$

#### Explanation:

We must recall our laws of radicals:

$\sqrt{a \cdot b \cdot c} = \sqrt{a} \cdot \sqrt{b} \cdot \sqrt{c}$

So the next thing to do is to find the prime factors of $145$

$145 = 5 \cdot 29$

$\implies \sqrt{145} = \sqrt{5} \cdot \sqrt{29}$

But we see $\sqrt{5}$ and $\sqrt{29}$ are both in the simplest form as they are both prime, cant be factored any more

Hence $\sqrt{145}$ is in its simplest form already

$\sqrt{145} \approx 12.042$