2
the integral of the square-root of (x-1)2 = x2/2 - x + C
-cos(x) + constant
4x^2 = 12 Divide through by '4' x^2 = 3 Square root both sides x = +/-sqrt(3)
(x3 + 4x2 - 3x - 12)/(x2 - 3) = x + 4(multiply x2 - 3 by x, and subtract the product from the dividend)1. x(x2 - 3) = x3 - 3x = x3 + 0x2 - 3x2. (x3 + 4x2 - 3x - 12) - (x3 + 0x2 - 3x) = x3 + 4x2 - 3x - 12 - x3 + 3x = 4x2 - 12(multiply x2 - 3 by 4, and subtract the product from 4x2 - 12)1. 4x(x - 3) = 4x2 - 12 = 4x2 - 122. (4x2 - 12) - (4x2 - 12) = 4x2 - 12 - 4x2 + 12 = 0(remainder)
better place to ask would be yahoo answers
In real numbers there is no square root of -4x2 However, when using complex numbers where √-1 = i, then: √-4x2 = 2xi If you were asking, for example, with regard to the difference of two squares, eg y2 - 4x2, then you are actually finding the square root of 4x2 which is 2x, giving: y2 - 4x2 = (y + 2x)(y - 2x)
the integral of the square-root of (x-1)2 = x2/2 - x + C
x = square root of 3.
(√4x^2)(√36x)=2x(6√x)=12x√x
square root x
This is an easy question : root 2, root 3, root 5. Any square root of an integer which is not integral itself is irrational.
(2x - i)(2x + i)where i is the square root of negative 1.x = 0.5i, -0.5i
replace square root o x with t.
Only if the term under the radical (square root sign) can be simplified to a rational expression. For example, √(4x2).
The square root of 81 is an integral number.
-cos(x) + constant
4x^2 = 12 Divide through by '4' x^2 = 3 Square root both sides x = +/-sqrt(3)