Question 157496
I'll start where you left off


{{{y(y-5)=x}}} Start with the given equation



{{{y^2-5y=x}}} Distribute



{{{y^2-5y-x=0}}} Subtract "x" from both sides



Notice how we have a quadratic equation in the form of {{{ay^2+by+c}}} where {{{a=1}}}, {{{b=-5}}}, and {{{c=-x}}}. So to solve for y, we must use the quadratic formula {{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}



{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} Start with the quadratic formula



{{{y = (-(-5) +- sqrt( (-5)^2-4*(1)*(-x) ))/(2*(1)) }}} Plug in {{{a=1}}}, {{{b=-5}}}, and {{{c=-x}}}




{{{y = (5 +- sqrt( 25+4x ))/(2) }}} Multiply and simplify



{{{y = (5 + sqrt( 25+4x ))/(2) }}} or {{{y = (5 - sqrt( 25+4x ))/(2) }}} Break up the "plus/minus"



So the inverse of {{{x(x-5)= y}}} is {{{y = (5 + sqrt( 25+4x ))/(2) }}} or {{{y = (5 - sqrt( 25+4x ))/(2) }}} 



Note: the inverse of {{{x(x-5)= y}}} is NOT a function.