# Thread: Solving for x in derivative = 0

1. ## Solving for x in derivative = 0

I have found the derivative of f(x) from a previous problem, and now I must set the derivative to 0 since I am now finding the function of x values where the tangent is horizontal, meaning that the slope is 0.

Here is where I am in the problem:

3x^2-8x-7=0

in an attempt to move x to one side and numbers to the other:

x^2-8x=7(1/3)

since x^2=2x

2x-8x=7(1/3)

But how do I get rid of the two x's since there is only supposed to be one x on the left side?

Sorry, I'm stuck!

2. Originally Posted by theredqueentheory

I have found the derivative of f(x) from a previous problem, and now I must set the derivative to 0 since I am now finding the function of x values where the tangent is horizontal, meaning that the slope is 0.

Here is where I am in the problem:

3x^2-8x-7=0

in an attempt to move x to one side and numbers to the other:

x^2-8x=7(1/3)

since x^2=2x

2x-8x=7(1/3)

But how do I get rid of the two x's since there is only supposed to be one x on the left side?

Sorry, I'm stuck!
$\displaystyle 3x^2-8x-7=0$

Use the quadratic formula: $\displaystyle x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

If, $\displaystyle x^2 = 2x$ (which is unusual) then $\displaystyle 2x-8x = \frac{7}{3}$ is equal to $\displaystyle -6x = \frac{7}{3}$