# How do you simplify sqrt42 rounded to the nearest tenths place?

Nov 19, 2016

$\sqrt{42} \approx 6.5$

#### Explanation:

If you ask a calculator to find $\sqrt{42}$ then it will give you an approximation like:

$\sqrt{42} \approx 6.4807406984$

So this is between $6.4$ and $6.5$, but much closer to $6.5$ than $6.4$.

So rounded to the nearest tenths place, $\sqrt{42} \approx 6.5$

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How would you find this without a calculator?

Note that:

$6 \cdot 6 = 36$

$6 \cdot 7 = 42$

$7 \cdot 7 = 49$

$42$ is nearly half way between ${6}^{2} = 36$ and ${7}^{2} = 49$

So we expect $\sqrt{42}$ to be about halfway between $6$ and $7$.

Looking at the graph of $y = {x}^{2}$ for $x$ between $6$ and $7$ and drawing a chord we see:

graph{(y-x^2)(y - (36+13(x-6))) = 0 [5.8, 7.2, 32, 51]}

By the time we get to $x \in \left[6 , 7\right]$ the parabola is pretty straight, so we expect this linear interpolation to be good.

In fact we find:

${6.5}^{2} = 42.25$

Using Newton's method, the error in the approximation will be about:

$\frac{0.25}{2 \cdot 6.5} = \frac{0.25}{13} \approx 0.02$

This is much smaller than $0.1$ so will not affect the tenths place.

Another way of looking at this is continued fractions.

We find that any number of the form $n \left(n + 1\right)$ has a square root given by:

sqrt(n(n+1)) = [n;bar(2,2n)] = n + 1/(2+1/(2n+1/(2+1/(2n+1/(2+...)))))

In our example $42 = 6 \left(6 + 1\right)$, so

$\sqrt{42} = 6 + \frac{1}{2 + \frac{1}{12 + \frac{1}{2 + \frac{1}{12 + \frac{1}{2 + \ldots}}}}}$

Truncating this, we can find rational approximations, such as:

sqrt(42) ~~ [6;2] = 6+1/2 = 6.5

sqrt(42) ~~ [6;2,12] = 6+1/(2+1/12) = 6+12/25 = 6.48