# finding square root

I want to know how to find the square root/cube root of a number ending with zero.

For instance, $\sqrt{1024}$ : though it's a perfect square, there is a shortcut. Like wise how can I find the $\sqrt{2240}$ or any number ending with zero?

• You punch 2, 2, 4, 0, sqrt on your calculator. – Gerry Myerson Aug 12 '16 at 7:16
• Maybe you can find a shortcut for numbers ending with two zeros. – N74 Aug 12 '16 at 7:24
• Is there a shortcut to calculate $\sqrt{1024}$ ? – user228113 Aug 12 '16 at 7:26
• By the way, it is true that $\sqrt{100\cdot x}=10\sqrt{x}$ and $\sqrt[3]{1000\cdot x}=10\sqrt[3]{x}$, but in the end $\sqrt{10},\sqrt[3]{10}$ and $\sqrt[3]{100}$ are irrational, so not much to do about them. – user228113 Aug 12 '16 at 7:27
• For $n>0,$ a computer may put $n=4^ab$ where $a\in \mathbb Z$ and $1/4\leq b<1,$ and compute $\sqrt n=2^a\sqrt b.$... A good first approximation for $\sqrt b$ is $(31b+5)/(14b+22),$ which is within about $0.4$% of $\sqrt b$ when $1/4\leq b <1.$ – DanielWainfleet Dec 24 '16 at 10:42

By $$(10a+b)^2 = a\times 100a+(10\times 2a+b)\times b$$

We can use long division,

\begin{array}{rrrr} & & 3 & 2 \\ & & -- & -- \\ 3 & \sqrt{} & 10 & 24 \\ & & 9 & \\ & & -- & -- \\ & & 1 & 24 \\ 62 & & 1 & 24 \\ & & -- & -- \\ & & & 0 \\ \end{array}

which means $32^2=(10\times 3+2)^2=100\times 9+62 \times 2=1024$,

thus $\sqrt{1024}=32$.

By $$(10a+b)^3 = a^2\times 1000a+(100\times 3a^2+10\times 3ab+b^2)\times b$$

We can use long division,

\begin{array}{rrr} & & 2 & 3 \\ & & -- & -- \\ 4 & \sqrt[3]{} & 12 & 167 \\ & & 8 & \\ & & -- & -- \\ & & 4 & 167 \\ 1389 & & 4 & 167 \\ & & -- & -- \\ & & & 0 \\ \end{array}

where $4=\color{red}{2}^2$ and $1389=100\times 3\times \color{red}{2}^2+10\times 3\times \color{red}{2} \times \color{blue}{3}+\color{blue}{3}^2$

therefore $\sqrt[3]{12167}=23$.

Only perfect squares will give you an integer answer when finding the square root and only perfect cubes will give you an integer answer when finding the cube root. For non-perfect squares/cubes you will always have an irrational (never ended non-repeating) decimal answer.

In generally you want to write the number as the product of whichever squares you can find as factors when finding a square root.

E.g. $1024$ divides by $4=2^2$ (a square) so write it as $1024=2^2\times256$. The number $256$ also divides by $4$ so write it as $1024=2^2\times2^2\times64$. Then we notice that $64$ is a square so we have: $1024=2^2\times2^2\times8^2=(2\times2\times8)^2=32^2$ and hence $\sqrt{1024}=\sqrt{32^2}=32$

For non-perfect squares you can do similar to reduce them down but only so far.

E.g. $2240$ divides by $4$ so write it as $2240=2^2\times505$. Examining the number $505$ notice that it is $5\times101$ which are both prime. So it can not be write in terms of further primes. So $\sqrt{2240}=\sqrt{2^2\times505}=2\sqrt{505}$ This could be approximated by a decimal value (easiest using a calculator) if needed.

Note that numbers ending in $00$ will divide by $100$ so you can write then as $10^2\times something$.

E.g. $57600=10^2\times576=10^2\times2^2\times144=10^2\times2^2\times12^2=(10\times2\times12)^2=240^2$ so $\sqrt{57600}=\sqrt{240^2}=240$.

For cube roots you want to do the same process but look for cubes which divide into the number. E.g. $1728$ divides by $8=2^3$ so $1728=2^3\times216$. Then number $216$ is equal to $6^3$ so $1728=2^3\times6^3=(2\times6)^3=12^3$. Hence $\sqrt[3]{1728}=\sqrt[3]{12^3}=12$

• @duanedibbley thanks for the edit. – Ian Miller Aug 12 '16 at 13:18

Find the prime factorisation of your original number, $$n=\prod_{i=1}^{k}p_i^{e_i}$$

with $p_i$ being distinct primes, $0<e_i\in \mathbb{N}$, etc.. Divide all exponents by 2. This is the factorisation of the square root, $$\sqrt{n}=\prod_{i=1}^{k}p_i^{e_i/2}$$

Take the product of the primes raised to the integer part of their new exponent. Call this product, $a$.

Take the product of one copy of each prime which has a non-integer new exponent. Call this second product, $b$.

Now $$\sqrt{n} =a\cdot \sqrt{b}$$

More generally, you can use this method to find cube, quartic, quintic, etc. roots in a similar way, dividing by that number, then grouping together primes with exponents equivalent modulo the level of root taken, such that $b'$ includes primes exponentiated sufficiently to account for the non-integer part of the new exponent, and placed inside an appropriate surd.