Not a "help need" but a question about "a self made square root formula"

[Angel11]

Hello,first time posting a thread not just here but generally so i'll try my best.
So while i was in class we were learning about square roots,at first it seemed fairly easy,but when i asked my math teacher how do we find them more easily, he smiled and talled me:"The problem is,you just don't".So that put me into a lot of thinking when i went home about if it is really impossible to simplify square roots at the point where i couldn't function properly.after about 2 days of thinking and trying diffrent ways i found out this one:If you divide your number (without the square rooting symbole) by 4 and then you square root the result with the square root of 4 then the result is the answer.For example if i want to find the \sqrt{36} then we do: 36/4=9. \sqrt{9}*\sqrt{4}=\sqrt{36} or \sqrt{9}*\sqrt{4}=6 which is the resolt of \sqrt{36}.My question to this is:has it ever been found becouse it seems fairly easier to do it that way instead of multipling numbers until you find it. Also i don't want to call it a solution becouse you still must find a square root

 

[MarkFL]

Hello and welcome to MHB, Angel1! (Wave)

There are algorithms, similar to long division, for manually computing square (and other) roots. I don't think they are taught much now though.

What you've discovered is a property of exponents...namely (let's assume all values are non-negative):

\(\displaystyle \left(a^b\cdot c^d\right)^n=a^{bn}\cdot c^{dn}\)

A square root is equivalent to an exponent of \(\displaystyle \frac{1}{2}\), that is:

\(\displaystyle \sqrt{x}=x^{\frac{1}{2}}\)

So, if you don't readily see that the radicand (the number under the radical) is a perfect square, but you do see it is the product of perfect squares, then you may state:

\(\displaystyle \sqrt{x}=\sqrt{a^2\cdot b^2}=\left(a^2\cdot b^2\right)^{\frac{1}{2}}=a^{2\cdot\frac{1}{2}}\cdot b^{2\cdot\frac{1}{2}}=a\cdot b=ab\)

You don't need to use rational exponents yet, as we can simply write:

\(\displaystyle \sqrt{x}=\sqrt{a^2\cdot b^2}=\sqrt{a^2}\cdot\sqrt{b^2}=a\cdot b=ab\)

 

[HallsofIvy]

Hello,first time posting a thread not just here but generally so i'll try my best.
So while i was in class we were learning about square roots,at first it seemed fairly easy,but when i asked my math teacher how do we find them more easily, he smiled and talled me:"The problem is,you just don't".So that put me into a lot of thinking when i went home about if it is really impossible to simplify square roots at the point where i couldn't function properly.after about 2 days of thinking and trying diffrent ways i found out this one:If you divide your number (without the square rooting symbole) by 4 and then you square root the result with the square root of 4 then the result is the answer.For example if i want to find the \sqrt{36} then we do: 36/4=9. \sqrt{9}*\sqrt{4}=\sqrt{36} or \sqrt{9}*\sqrt{4}=6 which is the resolt of \sqrt{36}.My question to this is:has it ever been found becouse it seems fairly easier to do it that way instead of multipling numbers until you find it. Also i don't want to call it a solution becouse you still must find a square root

That works as long as the square root is an integer.

If not, for example, if we want to find the square root of 30, we know that 25= 5^2 and 36= 6^2 are perfect squares, and 30 is between them. We also know that "squaring" is an increasing function (if a< b then a^2< b^2) so the square root of 30 must be between 5 and 6. We don't know where but the simplest "guess" would be half-way between 5 and 6, 5.5. (5.5)^2= 30.25. That is larger than 30 so we know the root must be between 5 and 5.5 so we might now try half way between those, 5.25. Repeat that until the two endpoints of the interval are "close enough" together, within whatever error is acceptable.

This method is referred to as "bisection" because at each step we halve the interval. Notice that 30 is a lot closer to 30.25 than to 25 so we might suspect that the root of 30 will be closer to 5.5 than to 5. We can make use of that by using the "secant" method. Instead of just choosing half way, we replace the curve y= sqrt(x) by the straight line from (5, 25) to (5.5, 30.25) then determine where that line crosses y= 30.