Question

Algebra 1A
help please thank you

Answer 1

x = 1 or x = 2 or x = 4

Explanation:

`\sqrt[x]{{441}^2} = \left({{441}^2}\right)^{^{(1)/(x)}}\\\\={\left(441\right)}^{(2)/(x)}\\\\={\left({21}^2\right)}^{(2)/(x)}\\\\={\left({21}\right)}^{(4)/(x)}`

`\text{For } {21}^{(4)/(x)} \text{ to be a whole number, } (4)/(x) \text{ must also be a whole number}`

⇒ x is a factor of 4

⇒ x = 1 or x = 2 or x = 4

Check:

When x = 1

`\sqrt[x]{{441}^2} = {\left({21}\right)}^{(4)/(x)}\\\\= {\left({21}\right)}^{(4)/(1)}\\\\= {\left({21}\right)}^(4)\\\\=194481 \text{ which is a whole number}`

When x = 2

`\sqrt[x]{{441}^2} = {\left({21}\right)}^{(4)/(x)}\\\\= {\left({21}\right)}^{(4)/(2)}\\\\= {\left({21}\right)}^(2)\\\\=441 \text{ which is a whole number}`

When x = 4

`\sqrt[x]{{441}^2} = {\left({21}\right)}^{(4)/(x)}\\\\= {\left({21}\right)}^{(4)/(4)}\\\\= {\left({21}\right)}^(1)\\\\=21 \text{ which is a whole number}`