How do you figure out the square root of a number?
In understanding square roots, it is helpful to start with squares. When we "square" a number, we multiply that number by itself; for example
`5^2 = 5` `times 5 = 25` or
`x^2 = x times x= x^2 ` or even
`(x+3)^2= (x+3)times (x+3) = (x^2 + 3x+3x+9) ` (remember to multiply both terms (the x and the 3) by each other and simplified `= x^2+6x+9`
Therefore, the square root is the number before it is multiplied by itself. It is the "inverse" operation for a square just as
`+` is the inverse of ` -`
and `times` is the inverse of `divide`
Finding a square root can be difficult when the number or expression is not solved easily. However, if we know the square root of the number before and after that number, we can estimate. For example, `7^2 = 49 therefore sqrt 49 =+-7` and `9^2=81 therefore sqrt 81 = +- 9`
Therefore, if we are looking for the square root of 64, (let's assume we do not know it), then we can see that the answer falls between 7 and 9 because 64 falls between 49 and 81.
Try it with numbers that are much harder to find the square root because the answer is a decimal; for example, `sqrt 5` . The number 5 lies between 4 (a perfect square because `2^2=4` ) and 9 ( a perfect square because `3^2=9` ). Now we can estimate that, as `sqrt 4=2` and `sqrt 9=3 ` , `sqrt 5 ` must be between 2 and 3, closer to 2 because 5 is much closer to 4 than it is to 9. We can then work steadily closer to our decimal answer if required through a method similar to long division, if an estimated answer is not sufficient.
Another way to work out a square root, especially useful for algebra is to use the exponent equivalent. An exponent is a power, just as the power in `x^2` is 2 or the power in `(2y)^4` is 4. The power for a square root is `1/2` which is the inverse of 2 (2 is effectively `2/1` and the inverse is `1/2` . For example, if we have to solve `sqrt x times x^2` this would be the same as saying `x^(1/2) times x^2 ` which, when applying the rules of exponents means we add the exponents if the bases are the same (in this case `x` is the base). Therefore, we get `x^(1/2) times x^2 = x^(1/2 + 2) = x^(5/2)`
Ans: When finding the square root of a number, either estimate and work from that estimate, or use the power method if calculators are not permitted or feasible.
There are a number of ways to extract square roots of positive integers without a calculator.
(1) Guess and refine -- estimate the root and correct; either your guess is too large or too small so add/subtract accordingly and try again. You can continue this process to as many decimal places as desired.
Ex: `sqrt(134) `
Since `11^2=121,12^2=144 ` we know the root is between 11 and 12. Try 11.5
`11.5^2=132.25 ` which is a little low. Next try 11.55
`11.55^2=133.4025 ` which is still a little low. Eventually you try 11.58 which is too big and you continue in this manner.
The obvious drawback is the large number of computations.
(2) A better method is known as the Babylonian method. This was taught in western schools until handheld calculators made the method obsolete. It is a special case of Newton's method of calculating roots from the calculus.
Again using `sqrt(134) ` we use an initial guess of 11.5
Then we iterate the following step: take your guess, a, and the next "guess" is computed by `1/2(a+134/a) ` .(Note that 134 is the number we want the root of.)
`1/2(11.5+134/11.5)=1065/92 ` which is approximately 11.57608696
Note that the approximations rapidly approach the value required. My calculator gives `sqrt(134)~~11.5758369 ` . Two iterations from the initial guess and the value is accurate to 7 decimal places.
There are other methods (see the link below) but the Babylonian method works very well.
To find the square root you have to find a number that when multiplied by it self makes the number your looking for
so the square root of 26 would be between 5 and 6 but its closer to 5 so you can say about 5
"Finding the square root of a number is the inverse operation of squaring that number." This means that the square root of a number is basically finding a number that can multiply by itself to get the number under your root. If you have to find the `sqrt(25)`, you could think "what number could I multiply by itself to get 25?" `5^(2)` or 5 x 5 gives you 25, meaning that the square root of 25 is 5. This is what is known as a perfect square because it is a whole number. Sometimes you will have decimals in you answer like with the `sqrt(5)`. For these types of problems, you can either put it into a calculator or estimate. If you know that `2^(2) = 4` and `3^(2) = 9`, you can estimate that the square root of 5 will be between 2 and 3. From there you can start with squaring 2.5 and go higher or lower as needed until you get an approximate answer.
When your not using a calculator, you can use the tree method. Say you want the square root of 24, you'd start like this, and divide until you get the smallest whole number with no remainders
3 * 8
2 * 4
2 * 2
Alright, now we are left with a 3, and three 2's. Take out any "pairs" of a number. In this case we have one pair of 2's. We take the pair out and leave in the other two numbers.
2 sqrt(3*2) = 2 sqrt(6)``
The rule to remember here is that only pairs make it out of the root.
There are also several ways you could have divided 24, but it would have still come to the same answer.
Lets consider the sqrt(180)
36 * 5
9 * 4
In this group we have two pairs, one pair of 3 and one pair of 2. Both come out of the square root like this
3 * 2 * sqrt(5)