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The root of a negative number is not real. Other people above showed it.
Be careful what you say; look.
In root(3-v) the v may be 3 at the most.
In general:
Root(expression) is only real for expression is not negative.

2006-07-07 19:02:54 · answer #1 · answered by Thermo 6 · 0 0

Domain is the set that has the values for which the function is defined. This is the basic definition of Domain but more precisely it is as follows

Given a function f:X→Y, the set X of input values is called the domain of f, and Y, the set of possible output values, is called the codomain. The range of f is the set of all actual outputs {f(x) : x in the domain}. Sometimes the codomain is incorrectly called the range because of a failure to distinguish between possible and actual values.

So when any number is squared it gives a positive result (i.e., the result is a positive number) and therefore contrarily the reverse also exists. So the domain of a square root function always x>=o.

2006-07-07 18:29:26 · answer #2 · answered by Sherlock Holmes 6 · 0 0

Because by its nature a square root function cannot be negative.

There is a branch of Mathematics that deals with negative square roots, but they are called imaginary numbers.

2006-07-07 18:23:36 · answer #3 · answered by rhutson 4 · 0 0

A + number times a + number is positive.
A - number times a - number is positive.
A - number times a + number is negative, but these are always different numbers, (-1 and 1 are completely different numbers) and the square root is asking "what number times itself equals the number underneath the funny looking square root roof?"

2006-07-07 18:39:04 · answer #4 · answered by Hume 1 · 0 0

You can't have a negative square root of a real number, so the solutions (y-positions) are all positive or 0.

2006-07-07 18:24:04 · answer #5 · answered by cptbirdman 2 · 0 0

until u come to the course of iota u cannot take out square root of negative number.

2006-07-07 18:30:06 · answer #6 · answered by sam_indian18 1 · 0 0

b/c negative numbers do not have square roots (at least in real numbers)

2006-07-07 18:24:14 · answer #7 · answered by Anonymous · 0 0

sqrt(x^2 + 7) has no regulations on x. The area is all actual numbers. we are able to take the sq. root of any efficient volume or 0. on the grounds that x^2 is continually efficient, x^2 +7 is continually efficient and there aren't any regulations on x.

2016-11-06 01:08:44 · answer #8 · answered by ? 4 · 0 0

There is no real number that is negative, and you can multiply it by itself to get a negative number.

Remember:

(-)(-) = (+) Always for real numbers.

2006-07-07 18:23:25 · answer #9 · answered by powhound 7 · 0 0

If -a * -a = +A and +b * +b = +B then ? * ? = - !

2006-07-07 18:29:30 · answer #10 · answered by temptnu37076 2 · 0 0

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