How can we find square root of a number by hand?

Monday, December 1, 2014

How can we find square root of a number by hand?

To be able to evaluate this, recall that with a square root of N such that $\displaystyle{N}={A}{B}$ , it follows square root of N lies in between the A and B.

A<sqrt(N)< B

This is our clue that we can apply factoring for the value inside the square root sign.

For a radical $\displaystyle{\sqrt[{{n}}]{{{x}}}}$ , the parts are called:

n = index

x= radicand or value inside the radical sign.

A square root has an index of 2 which has a radical sign root(2)(x) or sqrt(x).

Suppose we have sqrt( 30).

Apply factoring on the radicand: $\displaystyle{30}={5}\cdot{6}$ then we know that

$\displaystyle{5}\lt\sqrt{{{30}}}\lt{6}$

To solve it numerically, note that an average of two number $\displaystyle\frac{{{A}+{B}}}{{2}}$ will be in between A and B.

Then, $\displaystyle{A}\lt\frac{{{A}+{B}}}{{2}}\lt{B}$ .

Average value $\displaystyle=\frac{{{5}+{6}}}{{2}}=\frac{{11}}{{2}}$ or $\displaystyle{5.5}$

Apply the average value to the factoring of the radicand such that:

radicand= (A+B)/2 * radicand/((A+B)/2)

Divide radicand by the average value:

$\displaystyle\frac{\sqrt{{{x}}}}{{{\left(\frac{{{A}+{B}}}{{2}}\right)}}}=\frac{{30}}{{{\left(\frac{{11}}{{2}}\right)}}}$

$\displaystyle={30}\cdot{\left(\frac{{2}}{{11}}\right)}$

$\displaystyle=\frac{{60}}{{11}}$

Then, factoring of the radicand: $\displaystyle{30}=\frac{{11}}{{2}}\cdot\frac{{60}}{{11}}$

and it follows it square root will lie in between:

$\displaystyle\frac{{60}}{{11}}\lt\sqrt{{{30}}}\lt\frac{{11}}{{2}}$

Note: $\displaystyle\frac{{60}}{{11}}\lt\frac{{11}}{{2}}$ since $\displaystyle\frac{{60}}{{11}}\approx{5.455}$ and $\displaystyle\frac{{11}}{{2}}={5.5}$

Note that the boundary values is approximately same as "5.5" then we can estimate the value of the square root: $\displaystyle\sqrt{{{30}}}\approx{5.5}$

For more accurate estimation, repeat the same procedure with the new set of factors of the radicand:

$\displaystyle{30}=\frac{{11}}{{2}}\cdot\frac{{60}}{{11}}$

Then,

average value $\displaystyle=\frac{{\frac{{11}}{{2}}+\frac{{60}}{{11}}}}{{2}}=\frac{{241}}{{44}}$ or $\displaystyle{5.477}$ rounded off.

$\displaystyle\frac{\sqrt{{{x}}}}{{{\left(\frac{{{A}+{B}}}{{2}}\right)}}}=\frac{{30}}{{{\left(\frac{{241}}{{44}}\right)}}}=\frac{{1320}}{{241}}$

then new factoring: $\displaystyle{30}={\left(\frac{{241}}{{44}}\right)}\cdot{\left(\frac{{1320}}{{241}}\right)}$

new range will be: $\displaystyle{\left(\frac{{1320}}{{241}}\right)}<\sqrt{{{30}}}<{\left(\frac{{241}}{{44}}\right)}$

or $\displaystyle{5.47718}\lt\sqrt{{{30}}}\lt{5.477273}$

Note that the boundary values is approximately same as "5.4772" then we can estimate the value of the square root: $\displaystyle\sqrt{{{30}}}\approx{5.4772}$