Square Roots Calculation Tricks

[susanto3311]

hi all...

i have problem about square roots for fast calculation, like below sample :

is there fast calculation method not commonly/usually ways.

it's possible?

please, see my picture?

thanks in advance..

susanto

 

[susanto3311]

i find article solving square root calculation

In this article we are solving squares
1) number with last digit 5.
2) from 30 to 50.
3) from 50 to 80.
4) from 80 to 100.
5) from 100 to 130.

please see SQUARE ROOTS TRICKS - MATH SHORTCUT TRICKS ~ QUESTION PAPER

but my question or my problem is how do make for number " from 10 to 30", with quickly method

like sample :

12^2+23^2 = ...
9^2+27^2+15^2 = ...
etc...

 

[soroban]

Hello, susanto3311!

I have problem about square roots . . . You mean "squares".

$(1)\;21^2+13^2$

With the sum of two squares, there is no shortcut.

$21^2 +13^2 \;=\;441+169 \;=\;610$

$(2)\;25^2 - 14^2$

Recall the identity: $\:a^2-b^2 \:=\: (a-b)(a+b)$

$25^2-14^2 \;=\;(25-14)(25+14) \;=\;(11)(39) \;=\;429$

$(3)\;25^2 + 16^2 - 24^2$

$25^2-24^2 + 16^2 \;=\;(25-24)(25+24) + 16^2 \;=\;(1)(49) + 256 =\;305$

$(4)\;21^2 - 19^2 + 8^2$

$(21 - 19)(21+19) + 8^2\;=\;(2)(40)+64\;=\;144$

 

[susanto3311]

Hello, Soroban!..

do you have a formula for a^2+b^2 = ...

thanks again!.

 

[pickslides]

Soroban already answered this:

With the sum of two squares, there is no shortcut.

:D

 

[susanto3311]

hi guys...

how to fast identity last digit number to answer multiple choice question not overall calculation, just identity to get possibility right answer...

please see my picture...

 

[Nono713]

You can take the expression modulo 10 to investigate the last digit:

$$21^2 + 13^2 \bmod{10} = \left ( 21 \bmod{10} \right )^2 + \left ( 13 \bmod{10} \right )^2 = 1^2 + 3^2 \bmod{10} = 1 + 9 \bmod{10} = 10 \bmod{10} = 0$$

So the last digit is a zero (and you find that by considering only the last digit of all calculations involved, basically). It's an easy way to discard possible solutions, since the correct solution must end in a zero (but not all solutions that end in a zero are correct, of course). You can also try with different bases, not just 10, and the more bases you check the more possibilities you can eliminate, though base 10 is the simplest and easiest one.

 

[pickslides]

Or if you don't want to use modular arithmetic then

$21^2+13^2$

consider the last digit from each number $1^2=1$ and $3^2=9$ , $9+1 = 10$ , last number being a zero..

 

[susanto3311]

hello pickslides..

your advice is great!

is your trick apply for all squares number?

 

[pickslides]

hello pickslides..

your advice is great!

is your trick apply for all squares number?

You will find it does...

Same idea works when expanding binomials

 

[susanto3311]

You will find it does...

Same idea works when expanding binomials

hello pickslides..

the next challenge, how to find quickly / identify 2 last number from right not only 1 last number..

i really you have brilliant idea..

 

[susanto3311]

another problem for cubes number ...

23^3-16^3
42^3-8^3

do you have brilliant formula?

 

[Prometheus1]

another problem for cubes number ...

23^3-16^3
42^3-8^3

do you have brilliant formula?

\(\displaystyle x^3-y^3 = (x-y)(x^2+xy+y^2)
\)

 

[susanto3311]

Or if you don't want to use modular arithmetic then

$21^2+13^2$

consider the last digit from each number $1^2=1$ and $3^2=9$ , $9+1 = 10$ , last number being a zero..

hello guys...

it's possible with or using pickslides's method, maybe need modify it..

 

[Nono713]

It's literally the same principle, play with the numbers a bit and see what you get, practice.

 

[MarkFL]

hello guys...

it's possible with or using pickslides's method, maybe need modify it..

If you wish to find the last 2 digits resulting from the sum of the squares of 2 two-digit numbers, then we can develop a formula as follows:

If the two digits of the first number are $a$ and $b$ and the two digits of the second number are $c$ and $d$, then we may write:

\(\displaystyle (10a+b)^2+(10c+d)^2=100\left(a^2+c^2\right)+20(ab+cd)+\left(b^2+d^2\right)\)

The portion that will affect the last two digits is:

\(\displaystyle 20(ab+cd)+\left(b^2+d^2\right)\)

So, the last two digits of this number will have the same last two digits as the original sum. :D

 

[susanto3311]

If you wish to find the last 2 digits resulting from the sum of the squares of 2 two-digit numbers, then we can develop a formula as follows:

If the two digits of the first number are $a$ and $b$ and the two digits of the second number are $c$ and $d$, then we may write:

\(\displaystyle (10a+b)^2+(10c+d)^2=100\left(a^2+c^2\right)+20(ab+cd)+\left(b^2+d^2\right)\)

The portion that will affect the last two digits is:

\(\displaystyle 20(ab+cd)+\left(b^2+d^2\right)\)

So, the last two digits of this number will have the same last two digits as the original sum. :D

hi Mark, i confuse, could you explain to me with more simple example
like :

45^2+34^=...

 

[MarkFL]

hi Mark, i confuse, could you explain to me with more simple example
like :

45^2+34^=...

In this case we have:

\(\displaystyle a=4,\,b=5,\,c=3,\,d=4\)

And so we compute:

\(\displaystyle 20(ab+cd)+\left(b^2+d^2\right)=20(20+12)+(25+16)=681\)

Thus we know the last 2 digits of the original sum are $81$, and to check, we have:

\(\displaystyle 45^2+34^2=3181\)

 

[susanto3311]

hi Mark, very clear...

thanks a lot...