Question 92078
{{{3(x^2-2x)=27}}} Factor out the leading coefficient 3. This step is important since we want the {{{x^2}}} coefficient to be equal to 1.




Take half of the x coefficient -2 to get -1 (ie {{{-2/2=-1}}})

Now square -1 to get 1 (ie {{{(-1)^2=1}}})




{{{3(x^2-2x+1)=27}}} Add this result (1) to the expression {{{x^2-2x}}}  inside the parenthesis


{{{3(x^2-2x+1)=27+1(3)}}} Add 1(3) to the other side (remember we factored out a 3)


Now the left side is a complete square


{{{3(x-1)^2=27+1(3)}}} Factor {{{x^2-2x+1}}} into {{{(x-1)^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>)




{{{3(x-1)^2=27+3}}} Multiply 1 and 3 to get 3


{{{3(x-1)^2=30}}} Combine like terms


{{{(x-1)^2=10}}} Divide both sides by 3



{{{x-1=0+-sqrt(10)}}} Take the square root of both sides


{{{x=1+-sqrt(10)}}} Add 1 to both sides


So the expression breaks down to

{{{x=1+sqrt(10)}}} or {{{x=1-sqrt(10)}}}



So our answer is approximately

{{{x=4.16227766016838}}} or {{{x=-2.16227766016838}}}


Here is visual proof


{{{ graph( 500, 500, -10, 10, -10, 10, 3x^2-6x-27) }}} graph of {{{y=3x^2-6x-27}}}



When we use the root finder feature on a calculator, we would find that the x-intercepts are {{{x=4.16227766016838}}} and {{{x=-2.16227766016838}}}, so this verifies our answer.