Solve each equation by graphing.
39. $x^3-3 x^2-6 x+8=0$
40. $x^3-8 x^2+15 x=0$
Answer (2)
The solutions to x 3 − 3 x 2 − 6 x + 8 = 0 are found by graphing the function and identifying x-intercepts.
The x-intercepts are x = − 2 , 1 , 4 .
The solutions to x 3 − 8 x 2 + 15 x = 0 are found by factoring the equation.
The solutions are x = 0 , 3 , 5 , so the final answer is x = − 2 , 1 , 4 and x = 0 , 3 , 5 .
Explanation
Problem Analysis We are given two cubic equations to solve by graphing. Our goal is to find the x-intercepts of each equation, which represent the solutions.
Solving the First Equation For the first equation, x 3 − 3 x 2 − 6 x + 8 = 0 , we need to find the values of x where the function f ( x ) = x 3 − 3 x 2 − 6 x + 8 intersects the x-axis. By graphing this function, we can visually identify these points. The graph intersects the x-axis at x = − 2 , x = 1 , and x = 4 . Therefore, the solutions to the first equation are x = − 2 , 1 , 4 .
Solving the Second Equation For the second equation, x 3 − 8 x 2 + 15 x = 0 , we need to find the values of x where the function g ( x ) = x 3 − 8 x 2 + 15 x intersects the x-axis. We can factor this equation as x ( x 2 − 8 x + 15 ) = 0 , which further factors to x ( x − 3 ) ( x − 5 ) = 0 . This gives us the solutions x = 0 , x = 3 , and x = 5 .
Final Answer Therefore, the solutions to the first equation are x = − 2 , 1 , 4 , and the solutions to the second equation are x = 0 , 3 , 5 .
Examples
Cubic equations can model various real-world phenomena, such as the volume of a container or the trajectory of a projectile. For instance, if you're designing a box with specific volume requirements, you might end up with a cubic equation to determine the dimensions. Solving these equations helps engineers and designers find feasible solutions that meet the required specifications. Understanding the roots of these equations allows for optimization and informed decision-making in practical applications.
The solutions to the first equation x 3 − 3 x 2 − 6 x + 8 = 0 are x = − 2 , 1 , 4 , and the solutions to the second equation x 3 − 8 x 2 + 15 x = 0 are x = 0 , 3 , 5 .
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