Question

Determine a positive real root of this equation using EES: $$2 x^{3}-10 x^{0.5}-3 x=-3$$

Short answer

Question: Determine the positive real root of the equation $$2x^3 - 10\sqrt{x} - 3x = -3$$ using the Engineering Equation Solver (EES) method.

Step-by-step solution

Rewrite the equation for EES

In this step, we want to write the equation in a form that is suitable for EES. We can achieve this by moving all terms to one side of the equation, which will result in an equation equal to zero. We will do this by adding 3 to both sides of the equation: $$2x^3 - 10\sqrt{x} - 3x + 3 = 0$$

Use a numerical method to find the root

We can use the Newton-Raphson method to find the root of the equation. First, we need the derivative of our function, which is: $$f'(x) = 6x^2 - \frac{5}{\sqrt{x}} - 3$$ Now we choose an initial guess for the root, let's say, $$x_0 = 1$$. We can then apply the Newton-Raphson method, using the formula: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$ We will continue to iterate using this formula until the difference between consecutive values of $$x_n$$ is less than a specified tolerance or until the maximum number of iterations is reached. After performing the Newton-Raphson iterations, we obtain a root value.

Check that the root is positive

After obtaining the root value from the Newton-Raphson method, we need to check whether the root is positive. If the root is positive, it satisfies our requirement, and we have found the positive real root for the given equation. If the root turns out to be negative, we may need to choose a different initial guess and repeat the iterations. By following these steps and applying the EES method, the positive real root of the equation $$2x^3 - 10\sqrt{x} - 3x + 3 = 0$$ can be determined.