Question

Let f(x)= x^2-4x-c. Find a nonzero value of c such that f(c)=c.​

Answer 1

The nonzero value of c will be:

• 
  `c = 6`

Explanation:

Given the function

`f\left(x\right)=\:x^2-4x-c`

`f\left(c\right)=\:c^2-4c-c`

as

`f(c) = c`

so

`c=\:c^2-4c-c`

switching the sides

`c^2-4c-c=c`

subtract c from both sides

`c^2-4c-c-c=c-c`

`c^2-6c=0`

`c\left(c-6\right)=0`

Using the zero factor principle

`\:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)`

`c=0\quad \mathrm{or}\quad \:c-6=0`

so, the solutions to the quadratic equations are:

`c=0,\:c=6`

Therefore, a nonzero value of c will be:

• 
  `c = 6`