1 Answer

Kierchon · Answer 1 · 2023-03-26T19:15:35+0000

Hello!

• To solve the roots ,we must factor the number, and ,count how many times its prime factors are repeated,.

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• If they are repeated at least equal to the root index, we ,can take this value out of the root,.

Let's follow these steps below:

(a) the real third roots of 343

Let's factorize 343 below:

$\begin{gathered} 343\text 7 \\ 49|7 \\ 7|7 \\ 1 \end{gathered}$

So, 343 can be written as 7³ = 7 * 7 * 7.

And as the index of this root is 3 we can cancel this exponent with the root, look:

$\begin{gathered} \sqrt[3]{343}=\sqrt[3]{7^3}=7 \\ \\ \\ \end{gathered}$

I'll solve in the same way, first factorizing 1,024:

$\begin{gathered} 1024|2 \\ 512|2 \\ 256|2 \\ 128|2 \\ 64|2 \\ 32|2 \\ 16|2 \\ 8|2 \\ 4|2 \\ 2|2 \\ 1 \end{gathered}$

So, 1,024 can be written as 2^10.

But we can write it as 2^5 * 2^5. Doing the same step, we will have:

$\sqrt[5]{1,024}=\sqrt[5]{2^5\cdot2^5}=2\cdot2=4$

Let's factorize 25:

$\begin{gathered} 25|5 \\ 5|5 \\ 1 \end{gathered}$

So, 25 = 5².

Look as the exponent will be canceled with the index of the root:

$\sqrt[2]{25}=\sqrt[2]{5^2}=5$

Final Answers:

• (a), 7

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• (b), 4

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• (c) ,5