How do you solve y=4x^2-x^3 for x?

2 Answers
Feb 20, 2018

The zeroes are #0# and #4#.

Explanation:

I'm assuming that you mean how to find the zeroes of the function. First, factor out the #x^2# from both terms.

#4x^2-x^3=0#

#(x^2)(4-x)=0#

Now, set each of these factors equal to zero and see what #x# values you get from them:

#color(white){color(black)( (x^2=0,qquadqquad4-x=0), (sqrt(x^2)=sqrt0,qquadqquad-x=-4), (x=0,qquadqquadx=4)):}#

The results are #x=0# and #x=4#.

Feb 21, 2018

You don't. You can solve for the "roots" of the equation or find a "y" for a given "x". But "x" is a continuous variable.
The roots are #x = 0# and #x = 4#.

Explanation:

"Solving an equation means either generating a dependent value (y) from an independent value (x) or finding its "roots" - those values of "x" which generate a "y" value of zero (0). You cannot "solve" for an independent variable.

To find the roots, you can rearrange the expression:
#y = 4x^2 - x^3#; #y = 4x^2(1 - x/4)#

From the second expression you can see that a value of #x = 0# is a root, and also #x = 4# from #1-x/4 = 0#; #-x/4 = -1#; #x = 4#.