Question 102618


{{{x^2-10x+24=0}}} Start with the given equation



{{{x^2-10x=-24}}} Subtract 24 from both sides



Take half of the x coefficient -10 to get -5 (ie {{{-10/2=-5}}})

Now square -5 to get 25 (ie {{{(-5)^2=25}}})




{{{x^2-10x+25=-24+25}}} Add this result (25) to both sides. Now the expression {{{x^2-10x+25}}} is a perfect square trinomial.





{{{(x-5)^2=-24+25}}} Factor {{{x^2-10x+25}}} into {{{(x-5)^2}}}  (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




{{{(x-5)^2=1}}} Combine like terms on the right side


{{{x-5=0+-sqrt(1)}}} Take the square root of both sides


{{{x=5+-sqrt(1)}}} Add 5 to both sides to isolate x.


So the expression breaks down to

{{{x=5+sqrt(1)}}} or {{{x=5-sqrt(1)}}}



{{{x=5+1}}} or {{{x=5-1}}}    Take the square root of 1 to get 1



{{{x=6}}} or {{{x=4}}} Now combine like terms


So our answer is

{{{x=6}}} or {{{x=4}}}



Here is visual proof


{{{ graph( 500, 500, -10, 10, -10, 10, x^2-10x+24) }}} graph of {{{y=x^2-10x+24}}}


Here we can see that the x-intercepts are {{{x=6}}} and {{{x=4}}}, so this verifies our answer.