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Find positive root of equation x3 equals 2x plus 5?
x^3=2x+5 x^3-2x-5=0 Graph the equation or use a calculator. x= 2.09455148
How many complex roots does 6x4 plus x3 - 5 equals 0 have?
It has two complex roots.
When y is inversely proportial to square of x .K the constant is equals 120 so find x when y is equal to x?
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
How do you find zeros of the function f(x) x3 - 2x 21?
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.
How do you find a cubic equation with integral coefficients that has the roots 2 and 4 plus i?
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
