Question

Use linear interpolation between consecutive perfect squares to find find an approximation of each value. Round your answer to the nearest tenth. Using these rules how would I do this with `sqrt3`?

I'm having some trouble figuring out how to do this, any chance someone could give an explanation how to solve it?

Thanks

Answer 1

See below. The answer is `sqrt(3)~=1.7`.

Explanation:

[What is linear interpolation?]
Linear interpolation is a method of curve fitting with linear plynomials.
Suppose two points `A(x_0,y_0)` and `B(x_1,y_1)` are given. If `x_0<=x<=x_1`, the value `y` can be approximated by substituting `x` to the formula of the line `AB`, that is, `y-y_0=(y_1-y_0)/(x_1-x_0)(x-x_0)`.
https://en.wikipedia.org/wiki/Linear_interpolation

[Solve this question]
In this question, `sqrt(1)=1` and `sqrt(4)=2` can be used for linear interpolation.

The formula of the line between `A(1,1)` and `B(4,2)` is as follows.
`y-1=(2-1)/(4-1)(x-1)`
`y-1=1/3 (x-1)`
`y=1/3 x + 2/3`

Substitute `x=3` to the formula. The result is `y=5/3=1.66…~=1.7`

[This is your turn]
Then, how about trying to approximate `sqrt(3)` to the nearest hundredth using interpolation between `C(2.89, 1.7)` and `D(3.24,1.8)`?
The answer will be `sqrt(3)~=303/175=1.731…~=1.73`.