Math Problem Statement

\frac{1}{46500}\cdot\left(3.407x^3+110,365x^2+567x-1560\right) $$$$ =0

Solution

We are tasked with solving the following equation:

146500(3.407x3+110365x2+567x1560)=0\frac{1}{46500}\cdot\left(3.407x^3 + 110365x^2 + 567x - 1560\right) = 0

Step-by-Step Solution:

  1. First, multiply both sides of the equation by 4650046500 to eliminate the fraction:

3.407x3+110365x2+567x1560=03.407x^3 + 110365x^2 + 567x - 1560 = 0

  1. Now, we have a cubic equation in the form:

3.407x3+110365x2+567x1560=03.407x^3 + 110365x^2 + 567x - 1560 = 0

Solving the Cubic Equation:

To solve this cubic equation analytically or numerically, we can use numerical methods (such as Newton's method) or attempt to factor it if possible. Since the coefficients are quite large, the equation is more suitable for numerical methods. Let me calculate the approximate roots of this equation.

Give me a moment to compute the roots.The approximate roots of the cubic equation are:

  1. x132,393.6x_1 \approx -32,393.6
  2. x20.121x_2 \approx -0.121
  3. x30.116x_3 \approx 0.116

Thus, the equation has three real roots: two small roots close to zero and one large negative root.

Would you like more details about how these roots were computed, or do you have any questions?

Here are 5 related questions:

  1. How can we verify the roots of a cubic equation?
  2. What are some common methods to solve cubic equations?
  3. How would the nature of the roots change if the cubic term were different?
  4. What are the applications of cubic equations in physics or engineering?
  5. Can we solve this equation by factoring, or is numerical computation the best approach?

Tip: For cubic equations, numerical methods like Newton's method or synthetic division are often used when factoring is difficult.

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Math Problem Analysis

Mathematical Concepts

Algebra
Cubic Equations
Numerical Methods

Formulas

Cubic equation: ax^3 + bx^2 + cx + d = 0
Newton's Method for approximating roots

Theorems

Fundamental Theorem of Algebra

Suitable Grade Level

Grades 11-12