Solving x = x(8-2x): My vs. Provided Solution

[emergentecon]

Homework Statement

x = x(8-2x)

Homework Equations

x = x(8-2x)

The Attempt at a Solution

My Solution

x = x(8-2x)
1 = 8 - 2x
2x = 7
x = 7/2

Provided Solution

x = x(8-2x)
x = 8x - 2x^2
2x^2 -7x = 0
x(2x - 7) =
x = 0 / x = 7/2

Is my approach wrong?
Is x=0 always a possible answer, in addition to the found solution, so, should I have simply included x=0 in my solution?

Thanks!

 

[rock.freak667]

Yes you would need to include x=0.

As your solution divided throughout by x, in doing so you limited your final answer as x could not be equal to zero (you can't divide by zero so x could be anything but zero).

 

[Ray Vickson]

Homework Statement

x = x(8-2x)

Homework Equations

x = x(8-2x)

The Attempt at a Solution

My Solution

x = x(8-2x)
1 = 8 - 2x
2x = 7
x = 7/2

Provided Solution

x = x(8-2x)
x = 8x - 2x^2
2x^2 -7x = 0
x(2x - 7) =
x = 0 / x = 7/2

Is my approach wrong?
Is x=0 always a possible answer, in addition to the found solution, so, should I have simply included x=0 in my solution?

Thanks!

The red flag that should be a warning to you is that you are dividing both sides of an equation by a common unknown---x in this case. You should always check that you are not dividing by zero, as that is never, ever allowed. So, when you look at your original equation and before you start dividing, ask yourself: could I ever be dividing by 0? If the answer is yes, you can't do it. But, 0 is a perfectly legitimate solution: when x = 0 are the two sides of the equation equal to each other? The answer is yes.

After a while, this kind of checking will (or should) become second nature to you.

 

[Mark44]

Notice that in the provided solution, they expanded the right side, and then brought all terms to the other side. After factoring the left side, they were able to find both solutions.