Question 93105

{{{x^2-6x-40=0}}} Start with the given equation



{{{x^2-6x=40}}} Add 40 to both sides



Take half of the x coefficient -6 to get -3 (ie {{{-6/2=-3}}})

Now square -3 to get 9 (ie {{{(-3)^2=9}}})




{{{x^2-6x+9=40+9}}} Add this result (9) to both sides. Now the expression {{{x^2-6x+9}}} is a perfect square trinomial.





{{{(x-3)^2=40+9}}} Factor {{{x^2-6x+9}}} into {{{(x-3)^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-3)^2=49}}} Combine like terms on the right side


{{{x-3=0+-sqrt(49)}}} Take the square root of both sides


{{{x=3+-sqrt(49)}}} Add 3 to both sides to isolate x.


So the expression breaks down to

{{{x=3+sqrt(49)}}} or {{{x=3-sqrt(49)}}}



{{{x=3+7}}} or {{{x=3-7}}}    Take the square root of 49 to get 7



{{{x=10}}} or {{{x=-4}}} Now combine like terms


So our answer is

{{{x=10}}} or {{{x=-4}}}



Here is visual proof


{{{ graph( 500, 500, -10, 10, -10, 10, x^2-6x-40) }}} graph of {{{y=x^2-6x-40}}}


Here we can see that the x-intercepts are {{{x=10}}} and {{{x=-4}}}, so this verifies our answer.


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Answer:

So the answer is A)