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

1 Answer
yumean1119
Dec 25, 2017

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#.

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