Answer :
- Rearrange the equation: $18x^6 + 7x^2 - 27x = 0$.
- Factor out $x$: $x(18x^5 + 7x - 27) = 0$, so $x=0$ is one solution.
- Approximate the real root of $18x^5 + 7x - 27 = 0$ using the Newton-Raphson method.
- The solutions are $x = 0$ and $x \approx 1.019871118822465$, so the final answer is $\boxed{0, 1.019871118822465}$
### Explanation
1. Understanding the Problem
We are given the equation $27x = 18x^6 + 7x^2$. Our goal is to find all solutions for $x$.
2. Rearranging the Equation
First, we rearrange the equation to have all terms on one side: $18x^6 + 7x^2 - 27x = 0$.
3. Factoring out x
Next, we factor out the common factor $x$: $x(18x^5 + 7x - 27) = 0$. This tells us that one solution is $x = 0$.
4. Analyzing the Remaining Equation
Now, we need to solve the equation $18x^5 + 7x - 27 = 0$. Let $f(x) = 18x^5 + 7x - 27$.
5. Finding an Interval for a Root
We can observe that $f(0) = -27$ and $f(2) = 18(32) + 14 - 27 = 576 + 14 - 27 = 563 > 0$. This indicates that there is at least one real root between 0 and 2.
6. Approximating the Real Root
Using the Newton-Raphson method, we can approximate the real root. The formula for the Newton-Raphson method is $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$. In our case, $f(x) = 18x^5 + 7x - 27$, so $f'(x) = 90x^4 + 7$. Starting with an initial guess of $x_0 = 1.2$, we iterate to find a root. After several iterations, we find that $x \approx 1.019871118822465$.
7. Final Solutions
We can verify that this is indeed a root by plugging it back into the equation: $f(1.019871118822465) \approx 0$. Therefore, the solutions to the equation $27x = 18x^6 + 7x^2$ are $x = 0$ and $x \approx 1.019871118822465$.
### Examples
Understanding polynomial equations and their solutions is crucial in many fields, such as engineering, physics, and computer science. For example, when designing a bridge, engineers need to solve polynomial equations to determine the forces acting on the structure and ensure its stability. Similarly, in computer graphics, polynomial equations are used to model curves and surfaces, allowing for realistic rendering of 3D objects. By solving the equation, we find the points where the function equals zero, which can represent equilibrium states or critical points in a system.
- Factor out $x$: $x(18x^5 + 7x - 27) = 0$, so $x=0$ is one solution.
- Approximate the real root of $18x^5 + 7x - 27 = 0$ using the Newton-Raphson method.
- The solutions are $x = 0$ and $x \approx 1.019871118822465$, so the final answer is $\boxed{0, 1.019871118822465}$
### Explanation
1. Understanding the Problem
We are given the equation $27x = 18x^6 + 7x^2$. Our goal is to find all solutions for $x$.
2. Rearranging the Equation
First, we rearrange the equation to have all terms on one side: $18x^6 + 7x^2 - 27x = 0$.
3. Factoring out x
Next, we factor out the common factor $x$: $x(18x^5 + 7x - 27) = 0$. This tells us that one solution is $x = 0$.
4. Analyzing the Remaining Equation
Now, we need to solve the equation $18x^5 + 7x - 27 = 0$. Let $f(x) = 18x^5 + 7x - 27$.
5. Finding an Interval for a Root
We can observe that $f(0) = -27$ and $f(2) = 18(32) + 14 - 27 = 576 + 14 - 27 = 563 > 0$. This indicates that there is at least one real root between 0 and 2.
6. Approximating the Real Root
Using the Newton-Raphson method, we can approximate the real root. The formula for the Newton-Raphson method is $x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$. In our case, $f(x) = 18x^5 + 7x - 27$, so $f'(x) = 90x^4 + 7$. Starting with an initial guess of $x_0 = 1.2$, we iterate to find a root. After several iterations, we find that $x \approx 1.019871118822465$.
7. Final Solutions
We can verify that this is indeed a root by plugging it back into the equation: $f(1.019871118822465) \approx 0$. Therefore, the solutions to the equation $27x = 18x^6 + 7x^2$ are $x = 0$ and $x \approx 1.019871118822465$.
### Examples
Understanding polynomial equations and their solutions is crucial in many fields, such as engineering, physics, and computer science. For example, when designing a bridge, engineers need to solve polynomial equations to determine the forces acting on the structure and ensure its stability. Similarly, in computer graphics, polynomial equations are used to model curves and surfaces, allowing for realistic rendering of 3D objects. By solving the equation, we find the points where the function equals zero, which can represent equilibrium states or critical points in a system.
