Find the real solutions of the following equation by graphing.
x^3-4x^2+3x=0
To solve the equation x^3 - 4x^2 + 3x = 0 by graphing, we can plot the graph of the equation and find the x-values where the graph intersects the x-axis. These intersection points will represent the real solutions of the equation.
To start, let's rearrange the equation to have one side equal to zero:
x^3 - 4x^2 + 3x = 0
x(x^2 - 4x + 3) = 0
Now, we can factor the equation:
x(x - 3)(x - 1) = 0
This gives us three possible x-values that make the equation equal to zero: x = 0, x = 3, and x = 1.
Now, let's plot the graph of the equation by substituting different values of x and finding the corresponding y-values:
For x = -2: y = (-2)^3 - 4(-2)^2 + 3(-2) = -8 + 16 - 6 = 2
For x = -1: y = (-1)^3 - 4(-1)^2 + 3(-1) = -1 - 4 - 3 = -8
For x = 0: y = 0^3 - 4(0)^2 + 3(0) = 0
For x = 1: y = 1^3 - 4(1)^2 + 3(1) = 1 - 4 + 3 = 0
For x = 2: y = 2^3 - 4(2)^2 + 3(2) = 8 - 16 + 6 = -2
For x = 3: y = 3^3 - 4(3)^2 + 3(3) = 27 - 36 + 9 = 0
Plotting these points on a graph, we get:
(see attached image)
We can see that the graph intersects the x-axis at x = 0, x = 3, and x = 1. Therefore, the real solutions of the equation x^3 - 4x^2 + 3x = 0 are x = 0, x = 3, and x = 1.