The root of equation x ^ 3-2x-3x-6 = 0 on interval [- 2,4] must belong to interval The root of equation x ^ 3-2x-3x-6 = 0 on interval [- 2,4] must belong to interval () A.[-2,1] B.[5/2,4] C.[1,7/4] D.[7/4,5/2]

Is the equation written wrong?
x^3-2x^2-3x-6=0?
If the equation f (x) = x ^ 3-2x ^ 2-3x-6 = 0 has roots in [a, b]
Then f (a) * f (b)

1. | 2x-1 | = | 3x + 1 | 2. | 1 / 2x - 2 | - 3 = a would like to ask these two questions. How to deal with the absolute value in the equation?
1. | 2X-1 | = |3X +1 |
2. | 1/2x - 2 | -3 = a

1)2X-1=±(3x+1)
2X-1=3x+1 OR 2X-1=-3x-1
x=-2 or x=0
2)1/2x - 2=±(a+3)
1/2x - 2=a+3 1/2x - 2=-a-3
x=2(a+5) x=2(-1-a)

If a and B are the two roots of the equation 2x ^ 2 + 3x-1 = 0, we can make a quadratic equation with one variable, and the two roots are a + 1 / B and B + 1 / A

Let x ^ 2 + KX + T = 0 and a and B be two of the equations to get a + B = - 3 / 2, a * b = - 1 / 2 (a + 1) / B, (B + 1) / a be two of the equations x ^ 2 + KX + t = 0. (a + 1) / B + (B + 1) / a = (a ^ 2 + A + B ^ 2 + b) / AB = ((a + b) ^ 2-2ab + A + b) / AB = (9 / 4 + 1-3 / 2) / (- 1 / 2) = - 7 / 2 = - K, k = 7 / 2 (a + 1) / b * (B + 1

Solve the equation 2x ^ 2 + 3x-1 = 0 and find a quadratic equation of one variable
1. Make its two roots twice of the original equation
2. Make its two roots smaller than the original one

Let two of the original equations be X1 and X2, then X1 + x2 = - 3 / 2, x1x2 = - # - 189; 1. Let two of the new equations be m and N, M = 2x1, n = 2x2, then M + n = 2x1 + 2x2 = - 3, Mn = 4x1 · x2 = - 2. Therefore, the obtained equation is: X & # - 178; + 3x-2 = 02. Let two of the new equations be p and Q, P = x1-1, q = x2-1, then p + q = X1 +

The root of equation x ^ 3-2x-3x-6 = 0 on interval [- 2,4] must belong to interval The root of equation x ^ 3-2x-3x-6 = 0 on interval [- 2,4] must belong to interval () A.[-2,1] B.[5/2,4] C.[1,7/4] D.[7/4,5/2]