What is the $\sqrt{145}$ $sqrt145$ in simplest radical form?

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Answer 1 · 2017-12-19T21:22:03

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

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

Answer 2 · 2017-12-19T21:25:41

Simplest form = $\sqrt{145}$ $sqrt(145)$

$\approx 12.042$ $approx 12.042$

We must recall our laws of radicals:

$\sqrt{a \cdot b \cdot c} = \sqrt{a} \cdot \sqrt{b} \cdot \sqrt{c}$ $sqrt(a*b*c) = sqrt(a)*sqrt(b)*sqrt(c)$

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

$145 = 5 \cdot 29$ $145 = 5 * 29$

$\Rightarrow \sqrt{145} = \sqrt{5} \cdot \sqrt{29}$ $=> sqrt(145) = sqrt(5)*sqrt(29)$

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

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

$\sqrt{145} \approx 12.042$ $sqrt(145) approx 12.042$

What is the √145sqrt145 in simplest radical form?