Sure. √6 , the square root of 6, is approximately 2.44949. It's an irrational number --you can't get it as the ratio of any two integers -- but it's real. See Can_a_real_number_be_an_irrational_number
If by "radical" you mean "square root of", then yes. Both square roots of 25 are real numbers.
No, it is not.
The number word for the number 6 is "six".
Yes well at least that is what is what ive learnt
Nearly any number you can think of is a Real Number. So 8 is a real number.
It can be.
No it is not.
Yes.
If by "radical" you mean "square root of", then yes. Both square roots of 25 are real numbers.
Six is a real number. If you mean "six ninths", yes, that's a real number. If you mean "one ninth", that's also a real number.
negative
Odd
Not necessarily. If it is the same radical number, then the signs cancel out. Radical 5 times radical 5 equals 5. But if they are different, then you multiply the numbers and leave them under the radical sign. Example: radical 5 * radical 6 = radical 30
I'm not quite sure, but when the number inside the radical (square root sign) is negative, there is a no real-number solution.
A number under a radical sign is known as a radicand.
Such an equation has a total of six roots; the number of real roots must needs be even. Thus, depending on the specific equation, the number of real roots may be zero, two, four, or six.
The number under the radical sign (also known as the radical) is called the radican.