What is the simplest radical form of $\sqrt{160}$ $sqrt160$ ?

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What is the simplest radical form of $\sqrt{160}$ $sqrt160$ ?

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Answer 1 · 2016-08-05T21:40:04

$4 \sqrt{10}$ $4sqrt10$

Explanation:

Write 160 as the product of its prime factors, then we know what we are dealing with.

$\sqrt{160} = \sqrt{2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5} = \sqrt{2^{5} \times 5}$ $sqrt160 = sqrt(2xx2xx2xx2xx2xx2xx5) = sqrt(2^5 xx 5)$

= $\sqrt{2^{5} \times 5} = \sqrt{2^{4} \times 2 \times 5}$ $sqrt(2^5 xx 5) = sqrt(2^4 xx 2 xx 5)$

= $4 \sqrt{10}$ $4sqrt10$

Answer 2 · 2016-08-05T21:44:53

Radicals can be split by multiplication. It helps to be able to find perfect squares underneath the radicals during the factorization, and $16$ $16$ is a convenient perfect square.

If it helps, try going in steps of factoring out $2$ $2$ .

$\sqrt{160}$ $sqrt(160)$

$\sqrt{2 \cdot 80}$ $sqrt(2*80)$

$\sqrt{2 \cdot 2 \cdot 40}$ $sqrt(2*2*40)$

$\sqrt{2 \cdot 2 \cdot 2 \cdot 20}$ $sqrt(2*2*2*20)$

$\sqrt{2 \cdot 2 \cdot 2 \cdot 2 \cdot 10}$ $sqrt(2*2*2*2*10)$

$= \sqrt{16 \cdot 10}$ $= sqrt(16*10)$

$= \sqrt{16} \cdot \sqrt{10}$ $= sqrt(16)*sqrt(10)$

Since $\sqrt{16} = 4$ $sqrt(16) = 4$ , we end up with $4 \sqrt{10}$ $color(blue)(4sqrt10)$ .

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What is the simplest radical form of $\sqrt{160}$ $sqrt160$ ?

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What is the simplest radical form of $\sqrt{160}$ $sqrt160$ ?