If x2 is negative it will have a maximum value If x2 is positive it will have a minimum value
As long as the range of values for x stays within the set of real numbers, then:|x2| = x2As the square of a real number will always be positive anyway.If on the other hand, x belongs to the set of imaginary numbers, then:|x2| = -x2The reason for that is that in the case of imaginary numbers, x2 will give you a negative number. Its absolute value then would be the negative of that.And with complex numbers it's, well... complex:(ai + b)2 = -a2 + 2abi + b2So:|(ai + b)2| = a2 + 2abi + b2
No. For 0 < x < 2, 2x is larger.
Not necessarily. f(x) = -1-x2 is negative for any test value of x, it is asymptotically negative infinity, but it is NEVER zero.
-x times -x = x2 (negative multiplied with negative = positive) sqrt of x2 = x
If x2 is negative it will have a maximum value If x2 is positive it will have a minimum value
0
This equation is unsolvable as x2 cannot be a negative number as would be suggested by this equation. Any number, even a negative, multiplied by itself will always give a positive number but this equation leaves you with x2 = -48
As long as the range of values for x stays within the set of real numbers, then:|x2| = x2As the square of a real number will always be positive anyway.If on the other hand, x belongs to the set of imaginary numbers, then:|x2| = -x2The reason for that is that in the case of imaginary numbers, x2 will give you a negative number. Its absolute value then would be the negative of that.And with complex numbers it's, well... complex:(ai + b)2 = -a2 + 2abi + b2So:|(ai + b)2| = a2 + 2abi + b2
No. For 0 < x < 2, 2x is larger.
Not necessarily. f(x) = -1-x2 is negative for any test value of x, it is asymptotically negative infinity, but it is NEVER zero.
The answer would be 3x if 'X2' is '2x', but if 'X2' is x2, then the total answer would be x2 + x.
-x times -x = x2 (negative multiplied with negative = positive) sqrt of x2 = x
The Answer to x2-10x250 is negative 2500, also -2500.
no, its subtraction. if x2 is over x to the first power, the exponent at the higher value subtracts the x1. whether the x2 is above or below x1, x1 is always being subtrated from x2. X2 X(1) x __ OR __ = X or _ X(1) X2 1
It is a quadratic equation and the value of x is -7 or 6 x2 - 42 = (-x) x2 + x - 42 = 0 (x + 7) (x - 6) = 0
The LCM is the absolute value of x.