What is the sqaure root of -180x^8
I got 6x^4 i radical 5
me too
Usually the i comes last, though.
what do you mean it comes last
usually a complex number is written a+bi, and coefficients come before variables.
I know it is tempting to put the √5 last so it does not get confused with following numbers, but that's just orthography.
here, a is zero, so just bi:
6√5 x^4 i
To find the square root of -180x^8, we can break it down into two parts: the square root of the magnitude (absolute value) and the square root of the imaginary unit (i).
First, let's find the square root of the magnitude:
√(180x^8)
To simplify, we can break down 180x^8 into its prime factors.
180 = 2^2 * 3^2 * 5
x^8 = (x^4)^2
Therefore, √(180x^8) becomes:
√(2^2 * 3^2 * 5 * (x^4)^2)
√(2^2) * √(3^2) * √(5) * √((x^4)^2)
2 * 3 * x^4 * √5
6x^4√5
Now, the square root of the imaginary unit i:
√i
The square root of i can be represented as:
√i = (√2/2 + √2/2i)
Finally, combining the square root of the magnitude and the square root of the imaginary unit, we get:
6x^4√5 * (√2/2 + √2/2i)
Therefore, the square root of -180x^8 is 6x^4√5 * (√2/2 + √2/2i).