The function of the following program is to find the root of the equation 2x & sup3; - 4x & sup2; + 3x-6 = 0 by dichotomy, and the absolute error is not more than 0.0001 #include "stdio.h" float f(float) { return(2*x*x*x-4*x*x-3*x-6);} main() {float m=-100,n=90,r; r=(m+n)/2; while(_____________ *f(n)!=0) {if(f(r)*f(n)

The function of the following program is to find the root of the equation 2x & sup3; - 4x & sup2; + 3x-6 = 0 by dichotomy, and the absolute error is not more than 0.0001 #include "stdio.h" float f(float) { return(2*x*x*x-4*x*x-3*x-6);} main() {float m=-100,n=90,r; r=(m+n)/2; while(_____________ *f(n)!=0) {if(f(r)*f(n)

First empty: F (m)
Second empty: n = R
The third empty: r = (M + n) / 2

Use Newton iteration method to find the root of the following equation near 1.5. 2x ^ 3-4x ^ 2 + 3x-6 = 0

f(x)=2x^3-4x^2+3x-6
f'(x)=6x^2-8x+3
x(n+1)=xn-(2xn^3-4xn^2+3xn-6)/(6xn^2-8xn+3)
x1=1.5
x2=2.3333
x3=2.0610
x4=2.0026
x5=2.0000
x6=2.0000
So x = 2

If one root of equation 2x ^ 2 + 3x + 5m = 0 is greater than 1 and the other root is less than 1, then the value range of M is?
2X ^ 2 + 3x + 5m = 0 is a parabola with an opening (I understand this). From the graph of straight parabola, we can get 2x1 ^ 2 + 3 ^ 1 + 5m

You can know that only when x = 1 is negative, there will be one greater than zero and one less than zero. You only need this one

If the equation 2x ^ 2-3x + k = 0 has a real root, find the range of K

The real root is B ^ 2-4ac > = 0
9-4 * 2 * k > = 0
k