Question 948126

{{{f(x) = 2x^2 + 6x}}}

{{{f(x) = 2(x^2 + 3x)}}}

{{{f(x) = 2(x^2 + 3x+_)-2*_}}} ...recall {{{a^2+2ab+b^2=(a+b)^2}}}, so we have {{{a=1}}} and we need {{{b}}} and since we have that {{{2ab=3}}}, then {{{2*1*b=3}}}=>{{{2*b=3}}}{{{b=3/2}}}

now square it and add to

{{{f(x) = 2(x^2 + 3x+(3/2)^2)-2*(3/2)^2}}}...this {{{2*(3/2)^2}}} we are multiplying by {{{2}}} because we have {{{2}}} in front of parentheses 


{{{f(x) = 2(x^2 + 3x+(3/2)^2)-2*(9/4)}}}

{{{f(x) = 2(x^2 + 3x+(3/2)^2)-cross(2)*(9/cross(4)2)}}}

{{{f(x) = 2(x +3/2)^2-9/2}}}