Assuming you know that your number is a perfect square, the square root of an even number is even, and the square root of an odd number is odd.
If you mean square root when you say 'root'then... Yes, only the same number can be a square root and the factors of an even number must contain at least one even number So, if one of the roots is even then the other is the same number and is therefore even.
6 is a even number but not a square number. Any number that does not have a whole number as its square root is not a square number.
The square root of 200 is not a whole number, and therefore can neither be classified as odd or even.
Yes, but only if you count a root at the tangent as a double root.
No. The square root of -24 isn't even real, let alone rational because the square root of any negative number is going to be an imaginary number.
A composite even number which is 10
Only whole numbers are even or odd, and the square root of 75 is not a whole number.
No, it may be not even a rational number. Square root of 2 is 1,414213562... for example.
No, it is not even a real number. The square root of negative 5 is the square root of 5, times i.
Find the integers that have perfect squares on either side of your number. For example if your number was 23, the square root would be between 4 and 5. Since 23 is closer to 25 than 16, the square root would be closer to 5. Then try some out. 4.9 is too high, 4.7 is too low, 4.8 is pretty close. 4.795 is even closer.
I cannot see how odd or even will determine the number of digits in the square root. For example:100 is an even number, and its square root is 10 (2 digits)121 is an odd number, and its square root is 11 (2 digits)Maybe the question is not phrased properly for what the questioner is asking.
The square root of any negative number is not even a 'real' number. It's called 'imaginary', because there's no number you can multiply by itself to get a negative number. The square root of 101 is irrational.
Well . . there is not even such a thing as the square root of -6, unless you are working with non-real numbers. The square root of + 6, though, is not a whole number.
The square root of negative 64 isn't even a real number. In math and engineering, it's called an "imaginary number".
No, it is not even a real number. Instead, it is what is known as an imaginary number.
Assuming that you mean the nth. root: two - a negative and a positive root.
No. The square root of a negative number is not a real number, but an imaginary number, because no real number squared equals a negative number.
Yes. it is because it is in the 2 times tables and the 2 is an even number too so yes 4 is an even number
yes you can and get a positive # too.... unless the odd # is more than the even #
81 and 121 are two examples out of many more
Yes. The square of a whole number is always a whole number. For example, 3 squared is 9, so the square root of 9 is 3. What you never have, is the square root of a whole number being a fraction that is not a whole number. The square root of a whole number is either a whole number or an irrational number. For example, the square root of 2 is irrational, because there are no 2 whole numbers a and b such that a/b squared is 2. This is not terribly difficult to prove, but I have already said too much; I have answered your question.
It is not possible to have a root number of 2 yet be bigger than 50. (The only number with a root of '2' is 4. This is clearly not bigger than 50).