x^3=2x+5 x^3-2x-5=0 Graph the equation or use a calculator. x= 2.09455148
It has two complex roots.
sorry, it is wrong the answer is : y=k/x2(square) {when y = x we will write y as x and leave x2(square) alone. like this: x=120/x2(square) x multiply by x2(square) is x3(cube) so cube root of 120 = x so x is 4.93
That factors to (x + 3)(x^2 - 3x + 7)Setting those equal to zero leaves one real root (-3) and one complex root: one half times (three minus i times the square root of 19) where i is the square root of negative one.
If it has integral coefficients and 4+i is a root then its conjugate, 4-i must also be a root. So the equation is f(x) = (x-2)*(x-4-i)*(x-4+i) where each factor is x minus a root. Then multiply these out. = (x-2)*(x2 - 8x + 17) = x3 - 2x2 - 8x2 + 16x + 17x - 34 = x3 - 10x2 + 33x - 34
If: x3+1 = 65 Then: x3 = 65-1 And: x3 = 64 So: x = 4 by means of the cube root function on the calculator
x3.
sqrt(x3 + x2) = x*sqrt(x+1) and that cannot be simplified further.
IMPOSSIBLE
No.
y2=x3+3x2
The end part of the question does not seem to make sense. The equation has three real roots, a single root at x = -1 and a double root at x = 1
It is x = -5
No, it is not.
cube root of 216 = 6 > X= 6
Oh, isn't that just a happy little math problem we have here! To find the square root of x cubed, we can simply take the square root of x and then multiply it by itself three times. It's like painting a beautiful landscape, just take it one step at a time and enjoy the process.
(xn+2-1)/(x2-1)