Question

2. Find the real-number root. V1296 O 36 and 3 6 6 and -6

Answer 1

we have

`\sqrt[4]{1,296}`

Remember that

1,296=(36)^2

and 36=6^2

so

1,296=(6^2)^2=6^4

substitute

`\sqrt[4]{1,296}=\sqrt[4]{6^4}=6`

answer is 6

Part 2

we have

`\sqrt[3]{0.125b^3}`

we have that

0.125=125/1,000=(5/10)^3

substitute

`\sqrt[3]{0.125b^3}=\sqrt[3]{((5)/(10))^3b^3}=(5b)/(10)=0.5b`

answer is 0.5b