Square root of negative fifteen?

All the roots of any number, real or complex, may be found with a simple algorithm: e^((θ+2πk)i/n) a^(1/n) where a^(1/n) represents the principal nth root of a and k=0,1,2,...(n-1). For a negative real number, θ is π.

√-15 = √15 e^((π+2πk)i/2) = √15 e^(π/2(2k+1)i) {k = 0,1}
= √15 e^(π/2 i) and √15 e^(3π/2 i)
= i √15 and -i √15

That said, if otherwise unqualified, "the square root" of a number refers to the principal square root. In that case, the answer is i√15.

Answer: either ± i√15 if you are looking for all square roots or just i√15 if you're only looking for the principal square root.

sqrt(-15) = sqrt(-1) * sqrt(15)

The sqrt(-1) is imaginary and is abbreviated by the letter i (lower case i). It's called "imaginary" because there isn't a real number that, when multiplied by itself, will result in -1.

Since sqrt(15) is already in its final form, the final answer is i*sqrt(15).

sr = square root -3 sr X= -15 sr X = -15 divided by -3 sr X = 5 therefore x = 5 squared x = 25

√(-15)
= √(-1 * 15)
= √(-1) * √15
= √(i^2) * √15 (i = imaginary number)
= i * √15
= i√15

Imaginary number is i ² where i ² = - 1

Thus √ (- 15 ) = √ (15 i ² ) = ± i √15

The answer is a so-called imaginary number, and the value is 3.873i to three decimal places. The square root of 15 is not a rational value...By definition i x i = -1. 3x3=9 (too low) and 4x4=16 (too high). By picking successive values between 3 and 4, for example 3.5; we see 3.5x3.5=12.25 which is still too low. Splitting the difference again at 3.75, we come up with 3.75x3.75=14.0625. Without going through the entire process, hopefully you see the idea, but generally in practice the calculator simplifies a very laborious task. Before that tables and/or slide rules were employed to avoid reinventing the wheel. Hope this is some help! I am sure there also better methods available, in fact I recall knowing how to use calculus determine square roots more easily but haven't used it in so long I've forgotten it.

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NOTE:            i = √ -1

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√-15 = √ -1·15 = √-1 ·√15 = i√15


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Do you know how to find the normal square root of 15?

It is less than 4, because 4*4 = 16.

The value to 28 digits is 3.8729833462074168851792654...

Now what is that imaginary unit called "i"?

It is the square root of -1.

Then i * i = -1. (and -i * -i = +i*i = -1 also)

for the square root of -15 we have the square root of ( 15 * i*i)

then we factor this into the square root of (15) * sqrt(i*i)

But the sqrt(i*i) = +i (or -i)

So we have the square root of (15) * i (or sqrt(15) * (-i) )

this is 3.8729833462074168851792654* i and
-3.8729833462074168851792654*i

They are the two square roots of -15.

There are always two square roots of a number (even if the number is negative)

it is just the square root of -1 multiplied by square root of 15
square root of neagative one is represented by [i] so
i(sqrt15)

√(-15)
= √[(15)*(-1)]
= (√15)(√-1)
= (√15)i