Question 946692
if the number {{{x}}} whose square {{{x^2}}} is {{{24}}} more than five times the number {{{5x}}}, then we have

{{{x^2=5x+24}}}...solve for {{{x}}}

{{{x^2-5x-24=0}}}...factor completely

{{{x^2+3x-8x-24=0}}}

{{{(x^2+3x)-(8x+24)=0}}}

{{{x(x+3)-8(x+3)=0}}}

{{{(x-8)(x+3) = 0}}}

solutions:


if {{{(x-8) = 0}}}=> {{{highlight(x=8)}}}
if {{{(x+3) = 0}}} =>{{{highlight(x=-3)}}}

so, {{{highlight(x=8)}}} and {{{highlight(x=-3)}}} are your numbers