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the integral of the square-root of (x-1)2 = x2/2 - x + C
4x^2 = 12 Divide through by '4' x^2 = 3 Square root both sides x = +/-sqrt(3)
-cos(x) + constant
(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)
x = square root of 3.
the integral of the square-root of (x-1)2 = x2/2 - x + C
(√4x^2)(√36x)=2x(6√x)=12x√x
square root x
(2x - i)(2x + i)where i is the square root of negative 1.x = 0.5i, -0.5i
This is an easy question : root 2, root 3, root 5. Any square root of an integer which is not integral itself is irrational.
Only if the term under the radical (square root sign) can be simplified to a rational expression. For example, √(4x2).
4x^2 = 12 Divide through by '4' x^2 = 3 Square root both sides x = +/-sqrt(3)
replace square root o x with t.
The square root of 81 is an integral number.
-cos(x) + constant