Calculating Square root of decimal number manually.

Question

https://youtu.be/tRHLEWSUjrQ

In general, it will be difficult to compute the square root of a decimal number manually?

Examples :

50.73

71.21

156.45

Answer 1

$7^2 =49 < 50.73 <8^2 =64$ so $7< \sqrt{50.73} <8$.

So $\sqrt{50.73} \approx 7$.

If $\sqrt{50.73}= 7+x$ where $0 < x < 1$ then $(7+x)^2 = 49 + 14x + x^2 = 50.73$ so

$14x + x^2= 1.73$. Now $x$ is some "tenth" decimal so $x^2$ is a "hundredth" decimal so $14x$ overwhelms the $x^2$ so if we approximate we can ignore the $x^2$ for now.

$14x \approx 1.73$ so $x \approx \frac {1.73}{14}$. We are trying to round to the tenths place so $\frac {1.73}{14}=\frac 1{10} \frac {17.3}{14}\approx 1$. so $0.1 = \frac {1.4}{14} < x\approx \frac {1.73}{14} <\frac {2.8}{14}=0.2$ so $0.1 < x < 0.2$ and $7.1 < \sqrt{50.73} < 7.2$.

So $\sqrt{50.73}\approx 7.1$ And $7.1^2 = (7+ 0.1)^2 = 49 + 1.4 + 0.01=50.41$.

If $\sqrt{50.73} = 7.1 + y$ then $(7.1+y)^2 = 7.1^2 + 14.2y + y^2 = 50.41 + 14.2y + y^2 = 50.73$.

So $14.2y + y^2 = 0.32$. Now $y$ is "pretty small" int the "hundredths" so $y^2$ is very small, in the ten thousandths. So if we approximate we can ignore the $y^2$ and

$14.2y \approx 0.32$ and $y \approx \frac {0.32}{14.2}=\frac 1{100}\frac {32}{14.2}\approx \frac 2{100}\approx 0.2$ and $0.02 < y < 0.03$.

So $\sqrt{50.73}\approx 7.12$ And $7.12^2 = (7.1 + 0.02)^2 = 50.41+ 2\cdot 0.02\cdot 7.1 + 0.0004= 50.41 + 0.284+0.0004= 50.6944$.

And that's as close as we can get rounding to two decimal place as $7.13^2= (7.1 + 0.03)^2 = 7.1^2 + 0.06\cdot 7.1 + 0.0009= 50.41+0.426 + 0.0009 = 50.8369 > 50.73$

So $\sqrt{50.73}\approx 7.12$

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$8^2 = 64 < 71.21 < 81 = 9^2$ so $8< \sqrt{71.21} < 9$

$(8+x)^2 = 64+16x + x^2=71.21$ so $16x \approx 71.21-64 = 7.21$ and $x\approx \frac {7.21}{16}=0.1\times \frac {72.1}{16}$ so $0.4 < \frac {7.21}{16} < .5$. (Closer to $.5$ than to $0.4$.

$8.5^2 = 64 + 8 + 0.25=72.25$.

$(8.5-y)^2 = 74.25 - 17y + y^2 =71.21$ so $17y \approx 1.04$ and $y\approx 0.01 \frac {104}{17}\approx 0.06$ so $\sqrt{71.21}\approx 8.44$

etc.

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$12^2 < 156 < 13^2$

$(12 + x)^2 = 144 + 24x + x^2 =156.45$ so $24x \approx 12$ and $x \approx 0.5$.

$12.5^2 = 144 + 12 + 0.24 = 156.25$

$(12.5 + y)^2 = 156.25 + 25y + y^2 = 156.45$ so $25y\approx 0.2$ so $y \approx 0.008\approx 0.01$ to the nearest handredth.

So $\sqrt{156.45}\approx 12.51$.

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IN GENERAL

To find the $\sqrt W$ is a matter of simply finding $y.xxx^2 < W< y.xxw^2$ and to find the next decimal approximate $next decimal = \frac {W-y.xxx^2}{2\times y.xxx}$.

Example to fine $45.23$ to two decimals:

$6^2 < 45 < 7^2$ so: $6$.

$\frac {45-36}{12}=\frac 9{12}\approx 0.7$.

$6.7^2=36+4.2+0.49=40.69$

$\frac {45.23-40.69}{13.4}=\frac {4.54}{13.4}\approx 0.03$

So $\sqrt {45.23}\approx 6.73$.