How to Approximate Square Root of a Number

[ricardogarcia]

The guy is amazing. These lessons brighten my day and let me recall all of my college years of math.

[ComposedBySam]

For anyone wondering why this works, these are the first two terms of the binomial expansion.

[btgrk]

Bigger the number, it's relatively more difficult to find next number which is a perfect square. Also, error tolerance varies accordingly.

[b.a.dieudonne]

Wow I started looking at your videos to keep my mind sharp. I am now forwarding them to my children. Amazing - you are a fantastic teacher thank you 🙏🏽

[marytredinnick]

Im 60 and have BS in Elementary and Special Education.Now that Im retired I'm enjoying learning algebra. I soooo wish i could have had you as a teacher. You're so good at explaining each step. Thanks so much❤

[aaronaaron]

This is magnificent... the elegance of mathematics never stops marveling me.

[kingminato]

We can use the tailor's series, f(x) = ✓x
f'(x)= 1/✓x
For each number we must find the perfect square a which is closer ( x > a)
f(x) ~ f'(a) + 1/2 × 1/✓a × (x-a)
f(138) ~ f(121) + 1/2 × 1/11 × 17
f(138) ~11, 7
We get the first decimal of ✓x, for more precision check the tailor's series .

[YourAveragePlayr]

Really appreciate this guy popping up on my feed once a day, extremely helpful.

[brettkowalski]

This guy is bringing back my migraines I used to have when I was doing math competitions in middle school and high school.

[honeyartstudios]

These daily clips might help me overcome my math trauma

[thalesnemo]

So straight forward and simple !

My recollections are that the textbook method was too complex !

[krishnagarg]

Thank you sir for such helpful tricks😊😊

[JSSTyger]

I would like to offer a similar method. Let G = guess and E = error and we want to find the square root of C. C = (G+E)² = G²+2GE+E². With a small enough E value, E² will be close to zero. Our equation now becomes an approximation. C~G²+2GE and E~(C-G²)/(2G). Now that you have your approximate error, simply add it to G to get your final estimate. If you choose to get more accuracy, you can revise your guess. This method can also be extended to cube roots, etc by knowing the binomial expansion formula and eliminating the terms that have powers of E greater than 1.

[olerask]

In general f(x) ~ f(a) + f'(a) * (x-a). That is
1. sqrt(x) ~ sqrt(a) + (x-a)/(2*sqrt(a))
2. cuberoot(x) ~ cuberoot(a) + (x-a)/(3*(cuberoot(a))^2), etc.
Fx. cuberoot(75) ~ cuberoot(64) + (75-64)/(3*(cuberoot(64))^2) = 4 + 11/(3*4^2) ~ 4.2292, where cuberoot(75) ~ 4.2172.

[MartiniREAL]

This is the coolest thing ever. I have tried to figure out how to find an approximation of a square root and now I can.

[kenbihler]

finally ! thank you ! I've been doing bad approximations all my life !!!

[ChrisM]

Very well explained. If only we all remember that we can use this 'way of thinking' in many, many more cases in our lives.

[stephenhousman]

To those of you taking calculus this approximation is a first order Taylor series for the square root of x.

[villageidiot]

Love these lessons, great explanations.

[Kelsey-qhrh]

You taught this WAYYY better than my online math lessons