# `root(5)(20)` Use Newton's method to approximate the given number correct to eight decimal places.

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Approximate `x=root(5)(20)`  by taking.

`f(x)=x^5-20`

`f'(x)=5x^4`

`x_(n+1)=x_n-((x_n)^5-20)/(5(x_n)^4)`

since `root(5)(32)` =2 and 32 is reasonably close to 20 , we will take x_1=2 and approximate until they agree to eight decimal places.

`x_2=x_1-((x_1)^5-20)/(5*(x_1)^4)`

`x_2=2-(2^5-20)/(5*2^4)=1.85`

`x_3=x_2-((x_2)^5-20)/(5*(x_2)^4)`

`x_3=1.85-(1.85^5-20)/(5*1.85^4)~~1.821486137`

`x_4=x_3-((x_3)^5-20)/(5*(x_3)^4)`

`x_4=1.821486137-(1.821486137^5-20)/(5*(1.821486137)^4)`

`x_4~~1.820565136`

`x_5=x_4-((x_4)^5-20)/(5*(x_4)^4)`

`x_5=1.820565136-((1.820565136^5-20)/(5*1.820565136)^4)`

`x_5~~1.820564203`

`x_6=x_5-((x_5)^5-20)/(5*(x_5)^4)`

`x_6=1.820564203-((1.820564203^5-20)/(5*1.820564203^4))`

`x_6~~1.820564203`

So `root(5)(20)~~1.82056420`

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## Related Questions

scisser | Student

Newton's method says you first must pick an answer that's reasonably close. `1^5 = 1` , and `2^5 = 32` . 2 is closer, so we'll call `x(0) = 2` . Solving for the fifth root of 20 is equivalent to solving `x^5 = 20` . Newton's method reads:

`x(n+1) = x(n) - f(x(n))/(f'(x(n))) `

`f(x) = x^5 - 20 `
`f'(x) = 5x^4 `

`x(n+1) = x(n) - (x(n)^5 - 20)/(5x(n)^4) `
`x(n+1) = x(n) - x(n)/5 + 4/x(n)^4 `
`x(n+1) = (4/5)x(n) + 4/x(n)^4 `

As you run the series, this number will adjust until it hones in on the true value.

`x(0) = 2 `
`x(1) = 1.85 `
`x(2) = 1.821486137`
`x(3) = 1.820565136 `
`x(4) = 1.820564203 `
`x(5) = 1.820564203 `

`root(5)20 = 1.82056420 `