Hello!
• To solve the roots ,we must factor the number, and ,count how many times its prime factors are repeated,.
,
• 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:
[tex]\begin{gathered} 343\text{ | }7 \\ 49|7 \\ 7|7 \\ 1 \end{gathered}[/tex]
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:
[tex]\begin{gathered} \sqrt[3]{343}=\sqrt[3]{7^3}=7 \\ \\ \\ \end{gathered}[/tex]
(b) the real fifth roots of 1,024:
I'll solve in the same way, first factorizing 1,024:
[tex]\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}[/tex]
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:
[tex]\sqrt[5]{1,024}=\sqrt[5]{2^5\cdot2^5}=2\cdot2=4[/tex]
(c) the real square roots of 25:
Let's factorize 25:
[tex]\begin{gathered} 25|5 \\ 5|5 \\ 1 \end{gathered}[/tex]
So, 25 = 5².
Look as the exponent will be canceled with the index of the root:
[tex]\sqrt[2]{25}=\sqrt[2]{5^2}=5[/tex]
Final Answers:
• (a), 7
,
• (b), 4
,
• (c) ,5
