Can someone please help me with this problem? I can not figure it out. Thank you.
solve 4x + x(x-1)=0,
find the x-intercepts of
f(x)=4x+x(x-1)
The x(-axis) intercepts are the points on the curve where y = 0.
In this case, you can find them by solving
0 = 4x + x(x-1) = x^2 + 3x = x(x+3)
This equation is satisfied at x=0 and x=-3
The x-intercepts are therefore at
(x,y) = (-3,0) and (0,0).
This equation can be solved by factoring:
4x + x(x-1)=0
x(4+x-1)=0
x(x+3)=0
therefore
x=0 or x+3=0
Can you take it from here?
Of course, I'd be happy to help you with this problem!
To solve the equation 4x + x(x-1) = 0, we can start by expanding the expression on the left side:
4x + x^2 - x = 0
Combining like terms, we get:
x^2 + 3x = 0
Now, to find the x-intercepts of the function f(x) = 4x + x(x-1), we need to solve the equation x^2 + 3x = 0. This equation represents the points where the graph of the function intersects the x-axis.
To solve this quadratic equation, we can factor out an x:
x(x + 3) = 0
Setting each factor equal to zero, we have:
x = 0 or x + 3 = 0
The first solution, x = 0, represents one of the x-intercepts of the function.
For the second solution, x + 3 = 0, we need to isolate x:
x = -3
So, the x-intercepts of the function f(x) = 4x + x(x-1) are x = 0 and x = -3.
I hope this explanation helps you understand how to solve the problem! Let me know if there's anything else I can assist you with.