Question 1198731


The rule we use for completing the square is that if we add {{{(b/2)^2}}} to both sides of the equation in order to complete the square.

But if a general quadratic equation have a coefficient of {{{a}}} in front of {{{x^2}}}:

{{{ax^2 + bx + c = 0}}}

we just divide the whole equation by "{{{a}}}" first, then carry on:

{{{x^2 + (b/a)x + c/a = 0}}}



{{{4x^2-18x=3}}}............divide the whole equation by {{{4}}}

{{{x^2-(18/4)x=3/4}}}

{{{x^2-(9/2)x=3/4}}}............. the coefficient of the x term is {{{b=9/2}}}, so {{{(b/2)^2= ((9/2)/2)^2=(9/4)^2}}}, add to both sides 

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

{{{(x^2-9/4)^2=3/4+81/16}}}

{{{(x^2-9/4)^2=93/16}}}


so, you need to add {{{(9/4)^2}}} to both sides