How do you find a one-decimal place approximation for $\sqrt[4]{45}$ $root4 45$ ?

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How do you find a one-decimal place approximation for $\sqrt[4]{45}$ $root4 45$ ?

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Answer 1 · 2015-10-16T19:16:33

$\sqrt[4]{45} = 2.6$ $root(4)45=2.6$ by calculator

Explanation:

$\sqrt[4]{45} = 2.6$ $root(4)45=2.6$ by calculator

Answer 2 · 2015-10-16T20:40:12

Use rational approximations for $\sqrt{3}$ $sqrt(3)$ and $\sqrt{5}$ $sqrt(5)$ to find

$\sqrt[4]{45} \approx 2.6$ $root(4)(45) ~~ 2.6$

Explanation:

$\sqrt[4]{45} = \sqrt{\sqrt{45}} = \sqrt{\sqrt{3^{2} \cdot 5}} = \sqrt{3} \sqrt{\sqrt{5}}$ $root(4)(45) = sqrt(sqrt(45)) = sqrt(sqrt(3^2*5)) = sqrt(3) sqrt(sqrt(5))$

$\sqrt{3} \approx \frac{7}{4}$ $sqrt(3) ~~ 7/4$

$\sqrt{\sqrt{5}} \approx \sqrt{\frac{9}{4}} = \frac{3}{2}$ $sqrt(sqrt(5)) ~~ sqrt(9/4) = 3/2$

So

$\sqrt[4]{45} \approx \frac{7}{4} \cdot \frac{3}{2} = \frac{21}{8} = 2.625 \approx 2.6$ $root(4)(45) ~~ 7/4 * 3/2 = 21 / 8 = 2.625 ~~ 2.6$

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How do you find a one-decimal place approximation for $\sqrt[4]{45}$ $root4 45$ ?

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