Question

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

Answer 1

Answer:

The nonzero value of c will be:

• $c = 6$

Step-by-step 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:

• $c = 6$