How do I solve --> the Square root of a negative number?

Question

For example the Square root of negative 81?

Answer 1

√- 81 . . .No real number solution

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Imaginary number solution

√- 81 =

i√81 =

9i

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Answer 2

The square root of a negative number cannot be any number in the real number system (that is, the real number line). This is because a number has to be negative, zero, or positive, so when multiplied by itself you either get zero (if the original number was zero) or a positive number back (because two positive numbers multiplied together will give you a positive, as will two negative numbers multiplied together).

There is, however, the concept of "imaginary numbers". In this case, you let the variable "i" represent √-1. So √-81 = √-1√81 = 9i. A number like "9i" doesn't have any meaning related to numbers as we generally know them (real numbers). But the concept of imaginary numbers has a number of uses in mathematics, and strangely enough the real world too.

Answer 3

technically square roots of negative numbers do not exist since a square root is the same numer multiplied by itself and a negative times a negative is still positve
the suqare root of a negative is called a complex number, specifically an imaginary numer
take the square root as if it were postive and then put a lowercase i next to the number, the i denotes that it is imaginary
the square root of negative 81 would 9i

Answer 4

Square roots of negative numbers aren't in the set of real numbers; use complex numbers instead.

The two square roots of sqrt(-81) are 9i and -9i, where i is a complex number, whose square is -1. Complex numbers behave much like real numbers, provided that i^2 = -1.

More at the sources.

Answer 5

There would no answer because a postive # times a postive # is postive & a negitive # times a negitive # also equals positive. If the negitive sign was on the outside of the square root then whatever the answer is, you would make it negitive. For example {let ^ represent a square root symbol}, -^9= -3; ^-9= no answer. Its kind of like I-3I= 3 but if you had a negitive sign on the outside the number would end up being negitive; -I3I= -3. Or -I-3I= -3.

Answer 6

y = ?(x² + 4) do no longer forget which you would be able to't that the sq. root of the an make it x + 2, that must be incorrect. an extremely speedy way of thinking approximately this could be the say via fact that x² is in there, meaning that the two advantageous and damaging numbers will grow to be advantageous (ex: 2² = 4 and additionally (-2)² = 4). So via fact that all huge style which you plug in for x are going to be advantageous, meaning that each physique huge style below the unconventional would be advantageous, and in turn meaning that the area is all real numbers or in era notation: (-? , ?)

Answer 7

1) for Sqrt((-81)^2) write: x = Sqrt((-81)^2)

2) square both sides: x^2 = (-81)^2

3) then we have: x^2 - (-81)^2 = 0

4) graph y = x^2 - 6561 and see where it crosses the x-axis. You will
see it crosses at both 81 and -81.

Answer 8

usually you cannot take the square root of a negative number, but using imaginary numbers the answer is 9i because i is sqrt-1

Answer 9

in this case we resort to the complex number -i to come to our resuce.

according to this, i * i = -1
hence i = (-1) ^1/2

now (-81)^1/2 = (-1 * 81)^1/2

(a *b )^1/2 = (a)^1/2 * b^1/2

thus
(-1 * 81)^1/2 = (-1)^1/2 * (81)^1/2
= i*9
= 9i

Answer 10

sqrt (-81) = sqrt (81 x -1)
= sqrt 81 x sqrt -1
Here we introduce the imaginary term i which is equal to sqrt -1
therefore
sqrt (-81) = 9i
Note the j notation is also used

Answer 11

Factor out the minus sign!
The sqrt(-81) = sqrt(81) * sqrt(-1)

By definition, sqrt(-1) = i
So you end up with sqrt(81) * i

The answer is 9i
If you count it, another answer is -9i